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2022AIME I 真题及答案

Category: AMC美国数学竞赛, 国际竞赛 Date: 2022年4月28日下午4:44

2022AIME I 真题及答案

2022AIMEII真题及答案

Problem1

Quadratic polynomials $P(x)$ and $Q(x)$ have leading coefficients $2$ and $-2,$ respectively. The graphs of both polynomials pass through the two points $(16,54)$ and $(20,53).$ Find $P(0) + Q(0).$

Solution 1 (Linear Polynomials)

Let $R(x)=P(x)+Q(x).$ Since the $x^2$ -terms of $P(x)$ and $Q(x)$ cancel, we conclude that $R(x)$ is a linear polynomial.

Note that $begin{alignat*}{8} R(16) &= P(16)+Q(16) &&= 54+54 &&= 108, \ R(20) &= P(20)+Q(20) &&= 53+53 &&= 106, end{alignat*}$ so the slope of $R(x)$ is $frac{106-108}{20-16}=-frac12.$

It follows that the equation of $R(x)$ is $[R(x)=-frac12x+c]$ for some constant $c,$ and we wish to find $R(0)=c.$

We substitute $x=20$ into this equation to get $106=-frac12cdot20+c,$ from which $c=boxed{116}.$

~MRENTHUSIASM

Solution 2 (Quadratic Polynomials)

Let $begin{align*} P(x) &= 2x^2 + ax + b, \ Q(x) &= -2x^2 + cx + d, end{align*}$ for some constants $a,b,c$ and $d.$

We are given that $begin{alignat*}{8} P(16) &= &512 + 16a + b &= 54, hspace{20mm}&&(1) \ Q(16) &= &hspace{1mm}-512 + 16c + d &= 54, &&(2) \ P(20) &= &800 + 20a + b &= 53, &&(3) \ Q(20) &= &hspace{1mm}-800 + 20c + d &= 53, &&(4) end{alignat*}$ and we wish to find $[P(0)+Q(0)=b+d.]$ We need to cancel $a$ and $c.$ Since $operatorname{lcm}(16,20)=80,$ we subtract $4cdot[(3)+(4)]$ from $5cdot[(1)+(2)]$ to get $[b+d=5cdot(54+54)-4cdot(53+53)=boxed{116}.]$ ~MRENTHUSIASM

Solution 3 (Pure Brute Force)

Let $[P(x) = 2x^2 + bx + c]$ $[Q(x) = -2x^2 + dx + e]$

By substitutes $(16, 54)$ and $(20, 53)$ into these equations, we can get: $[2(16)^2 + 16b + c = 54]$ $[2(20)^2 + 20b + c = 53]$ Hence, $b = -72.25$ and $c = 698$ .

Similarly, $[-2(16)^2 + 16d + e = 54]$ $[-2(20)^2 + 20d + e = 53]$ Hence, $d = 71.75$ and $e = -582$ .

Notice that $c = P(0)$ and $d = Q(0)$ . Therefore $P(0) + Q(0) = 698 + -582 = boxed{116}$

~Littlemouse

Problem2

Find the three-digit positive integer $underline{a},underline{b},underline{c}$ whose representation in base nine is $underline{b},underline{c},underline{a}_{,text{nine}},$ where $a,$ $b,$ and $c$ are (not necessarily distinct) digits.

Solution 1

We are given that $[100a + 10b + c = 81b + 9c + a,]$ which rearranges to $[99a = 71b + 8c.]$ Taking both sides modulo $71,$ we have $begin{align*} 28a &equiv 8c pmod{71} \ 7a &equiv 2c pmod{71}. end{align*}$ The only solution occurs at $(a,c)=(2,7),$ from which $b=2.$

Therefore, the requested three-digit positive integer is $underline{a},underline{b},underline{c}=boxed{227}.$

~MRENTHUSIASM

Solution 2

As shown in Solution 1, we get $99a = 71b+8c$ .

Note that $99$ and $71$ are large numbers comparatively to $8$ , so we hypothesize that $a$ and $b$ are equal and $8c$ fills the gap between them. The difference between $99$ and $71$ is $28$ , which is a multiple of $4$ . So, if we multiply this by $2$ , it will be a multiple of $8$ and thus the gap can be filled. Therefore, the only solution is $(a,b,c)=(2,2,7)$ , and the answer is $underline{a},underline{b},underline{c}=boxed{227}$ .

~KingRavi

Solution 3

As shown in Solution 1, we get $99a = 71b+8c.$

We list a few multiples of $99$ out: $[99,198,297,396.]$ Of course, $99$ can't be made of just $8$ 's. If we use one $71$ , we get a remainder of $28$ , which can't be made of $8$ 's either. So $99$ doesn't work. $198$ can't be made up of just $8$ 's. If we use one $71$ , we get a remainder of $127$ , which can't be made of $8$ 's. If we use two $71$ 's, we get a remainder of $56$ , which can be made of $8$ 's. Therefore we get $99cdot2=71cdot2+8cdot7$ so $a=2,b=2,$ and $c=7$ . Plugging this back into the original problem shows that this answer is indeed correct. Therefore, $underline{a},underline{b},underline{c}=boxed{227}.$

~Technodoggo

Solution 4

As shown in Solution 1, we get $99a = 71b+8c$ .

We can see that $99$ is $28$ larger than $71$ , and we have an $8c$ . We can clearly see that $56$ is a multiple of $8$ , and any larger than $56$ would result in $c$ being larger than $9$ . Therefore, our only solution is $a = 2, b = 2, c = 7$ . Our answer is $underline{a},underline{b},underline{c}=boxed{227}$ .

~Arcticturn

Problem3

In isosceles trapezoid $ABCD$ , parallel bases $overline{AB}$ and $overline{CD}$ have lengths $500$ and $650$ , respectively, and $AD=BC=333$ . The angle bisectors of $angle{A}$ and $angle{D}$ meet at $P$ , and the angle bisectors of $angle{B}$ and $angle{C}$ meet at $Q$ . Find $PQ$ .

Diagram

$[asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoint(A,D,C); P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D); Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C); draw(anglemark(P,A,B,1000),red); draw(anglemark(D,A,P,1000),red); draw(anglemark(A,B,Q,1000),red); draw(anglemark(Q,B,C,1000),red); draw(anglemark(P,D,A,1000),red); draw(anglemark(C,D,P,1000),red); draw(anglemark(Q,C,D,1000),red); draw(anglemark(B,C,Q,1000),red); add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$P$",P,1.5*NE,linewidth(4)); dot("$Q$",Q,1.5*NW,linewidth(4)); draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q); label("$500$",midpoint(A--B),1.25N); label("$650$",midpoint(C--D),1.25S); label("$333$",midpoint(A--D),1.25W); label("$333$",midpoint(B--C),1.25E); [/asy]$ ~MRENTHUSIASM ~ihatemath123

Solution 1

We have the following diagram: $[asy] /* Made by MRENTHUSIASM , modified by Cytronical */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, X, Y, Z, W; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoint(A,D,C); P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D); Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C); X = intersectionpoint(B--5*(Q-B)+B,C--D); Y = (0,6*sqrt(731)); Z = intersectionpoint(A--4*(P-A)+A,B--4*(Q-B)+B); W = intersectionpoint(A--5*(P-A)+A,C--D); draw(anglemark(P,A,B,1000),red); draw(anglemark(D,A,P,1000),red); draw(anglemark(A,B,Q,1000),red); draw(anglemark(Q,B,C,1000),red); draw(anglemark(P,D,A,1000),red); draw(anglemark(C,D,P,1000),red); draw(anglemark(Q,C,D,1000),red); draw(anglemark(B,C,Q,1000),red); add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$P$",P,1.5*(-1,0),linewidth(4)); dot("$Q$",Q,1.5*E,linewidth(4)); dot("$X$",X,1.5*dir(-105),linewidth(4)); dot("$Y$",Y,1.5*N,linewidth(4)); dot("$Z$",Z,4.5*dir(75),linewidth(4)); dot("$W$",W,1.5*dir(-75),linewidth(4)); draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q); draw(P--W^^Q--X^^Y--Z,dashed); [/asy]$

Let $X$ and $W$ be the points where $AP$ and $BQ$ extend to meet $CD$ , and $YZ$ be the height of $triangle AZB$ . As proven in Solution 2, triangles $APD$ and $DPW$ are congruent right triangles. Therefore, $AD = DW = 333$ . We can apply this logic to triangles $BCQ$ and $XCQ$ as well, giving us $BC = CX = 333$ . Since $CD = 650$ , $XW = DW + CX - CD = 16$ .

Additionally, we can see that $triangle XZW$ is similar to $triangle PQZ$ and $triangle AZB$ . We know that $frac{XW}{AB} = frac{16}{500}$ . So, we can say that the height of the triangle $AZB$ is $500u$ while the height of the triangle $XZW$ is $16u$ . After that, we can figure out the distance from $Y$ to $PQ: frac{500+16}{2} = 258u$ and the height of triangle $PZQ: 500-258 = 242u$ .

Finally, since the ratio between the height of $PZQ$ to the height of $AZB$ is $242:500$ and $AB$ is $500$ , $PQ = boxed{242}.$

~Cytronical

Solution 2

Extend line $PQ$ to meet $AD$ at $P'$ and $BC$ at $Q'$ . The diagram looks like this: $[asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, P1, Q1; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoint(A,D,C); P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D); Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C); P1 = intersectionpoint(A--D,P--(-300)*(Q-P)+P); Q1 = intersectionpoint(B--C,Q--300*(Q-P)+Q); draw(anglemark(P,A,B,1000),red); draw(anglemark(D,A,P,1000),red); draw(anglemark(A,B,Q,1000),red); draw(anglemark(Q,B,C,1000),red); draw(anglemark(P,D,A,1000),red); draw(anglemark(C,D,P,1000),red); draw(anglemark(Q,C,D,1000),red); draw(anglemark(B,C,Q,1000),red); add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$P$",P,1.5*NE,linewidth(4)); dot("$Q$",Q,1.5*NW,linewidth(4)); dot("$P'$",P1,1.5*W,linewidth(4)); dot("$Q'$",Q1,1.5*E,linewidth(4)); draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q); draw(P--P1^^Q--Q1,dashed); [/asy]$ Because the trapezoid is isosceles, by symmetry $PQ$ is parallel to $AB$ and $BC$ . Therefore, $angle PAB cong angle APP'$ by interior angles and $angle PAB cong angle PAD$ by the problem statement. Thus, $triangle P'AP$ is isosceles with $P'P = P'A$ . By symmetry, $P'DP$ is also isosceles, and thus $P'A = frac{AD}{2}$ . Similarly, the same thing is happening on the right side of the trapezoid, and thus $P'Q'$ is the midline of the trapezoid. Then, $PQ = P'Q' - (P'P + Q'Q)$ .

Since $P'P = P'A = frac{AD}{2}, Q'Q = Q'B = frac{BC}{2}$ and $AD = BC = 333$ , we have $P'P + Q'Q = frac{333}{2} + frac{333}{2} = 333$ . The length of the midline of a trapezoid is the average of their bases, so $P'Q' = frac{500+650}{2} = 575$ . Finally, $PQ = 575 - 333 = boxed{242}$ .

~KingRavi

Solution 3

We have the following diagram: $[asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, X, Y, Z, W; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoint(A,D,C); P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D); Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C); X = intersectionpoint(B--5*(Q-B)+B,C--D); Y = intersectionpoint(C--5*(Q-C)+C,A--B); Z = intersectionpoint(D--5*(P-D)+D,A--B); W = intersectionpoint(A--5*(P-A)+A,C--D); draw(anglemark(P,A,B,1000),red); draw(anglemark(D,A,P,1000),red); draw(anglemark(A,B,Q,1000),red); draw(anglemark(Q,B,C,1000),red); draw(anglemark(P,D,A,1000),red); draw(anglemark(C,D,P,1000),red); draw(anglemark(Q,C,D,1000),red); draw(anglemark(B,C,Q,1000),red); add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$P$",P,1.5*(-1,0),linewidth(4)); dot("$Q$",Q,1.5*E,linewidth(4)); dot("$X$",X,1.5*dir(-105),linewidth(4)); dot("$Y$",Y,1.5*N,linewidth(4)); dot("$Z$",Z,1.5*N,linewidth(4)); dot("$W$",W,1.5*dir(-75),linewidth(4)); draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q); draw(P--Z^^P--W^^Q--X^^Q--Y,dashed); [/asy]$ Extend lines $AP$ and $BQ$ to meet line $DC$ at points $W$ and $X$ , respectively, and extend lines $DP$ and $CQ$ to meet $AB$ at points $Z$ and $Y$ , respectively.

Claim: quadrilaterals $AZWD$ and $BYXD$ are rhombuses.

Proof: Since $angle DAB + angle ADC = 180^{circ}$ , $angle ADP + angle PAD = 90^{circ}$ . Therefore, triangles $APD$ , $APZ$ , $DPW$ and $PZW$ are all right triangles. By SAA congruence, the first three triangles are congruent; by SAS congruence, $triangle PZW$ is congruent to the other three. Therefore, $AD = DW = WZ = AZ$ , so $AZWD$ is a rhombus. By symmetry, $BYXC$ is also a rhombus.

Extend line $PQ$ to meet $overline{AD}$ and $overline{BC}$ at $R$ and $S$ , respectively. Because of rhombus properties, $RP = QS = frac{333}{2}$ . Also, by rhombus properties, $R$ and $S$ are the midpoints of segments $AD$ and $BC$ , respectively; therefore, by trapezoid properties, $RS = frac{AB + CD}{2} = 575$ . Finally, $PQ = RS - RP - QS = boxed{242}$ .

~ihatemath123

Solution 4

$[asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, X, Y, Z, W; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoint(A,D,C); P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D); Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C); draw(anglemark(P,A,B,1000),red); draw(anglemark(D,A,P,1000),red); draw(anglemark(A,B,Q,1000),red); draw(anglemark(Q,B,C,1000),red); draw(anglemark(P,D,A,1000),red); draw(anglemark(C,D,P,1000),red); draw(anglemark(Q,C,D,1000),red); draw(anglemark(B,C,Q,1000),red); add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$P$",P,1.5*NE,linewidth(4)); dot("$Q$",Q,1.5*NW,linewidth(4)); draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q); X = (-121,6*sqrt(731)); Y = (121,6*sqrt(731)); Z = (-121,-6*sqrt(731)); W = (121,-6*sqrt(731)); draw(X--Z^^Y--W,dashed); draw(rightanglemark(A,X,Z,500),red); draw(rightanglemark(B,Y,W,500),red); draw(rightanglemark(C,W,Y,500),red); draw(rightanglemark(D,Z,X,500),red); dot("$X$",X,1.5*N,linewidth(4)); dot("$Y$",Y,1.5*N,linewidth(4)); dot("$Z$",Z,1.5*S,linewidth(4)); dot("$W$",W,1.5*S,linewidth(4)); [/asy]$

Let $X$ and $Y$ be the feet of the altitudes from $P$ and $Q$ , respectively, to $AB$ , and let $Z$ and $W$ be the feet of the altitudes from $P$ and $Q$ , respectively, to $CD$ . Side $AB$ is parallel to side $CD$ , so $XYWZ$ is a rectangle with width $PQ$ . Furthermore, because $CD - AB = 650-500 = 150$ and trapezoid $ABCD$ is isosceles, $WC - YB = ZD - XA = 75$ .

Also because $ABCD$ is isosceles, $angle ABC + angle BCD$ is half the total sum of angles in $ABCD$ , or $180^{circ}$ . Since $BQ$ and $CQ$ bisect $angle ABC$ and $angle BCD$ , respectively, we have $angle QBC + angle QCB = 90^{circ}$ , so $angle BQC = 90^{circ}$ .

Letting $BQ = 333k$ , applying Pythagoras to $triangle BQC$ yields $QC = 333sqrt{1-k^2}$ . We then proceed using similar triangles: $angle BYQ = angle BQC = 90^{circ}$ and $angle YBQ = angle QBC$ , so by AA similarity $YB = 333k^2$ . Likewise, $angle CWQ = angle BQC = 90^{circ}$ and $angle WCQ = angle QCB$ , so by AA similarity $WC = 333(1 - k^2)$ . Thus $WC + YB = 333$ .

Adding our two equations for $WC$ and $YB$ gives $2WC = 75 + 333 = 408$ . Therefore, the answer is $PQ = ZW = CD - 2WC = 650 - 408 = boxed{242}$ .

~Orange_Quail_9

Solution 5

This will be my first solution on AoPS. My apologies in advance for any errors.

Angle bisectors can be thought of as the locus of all points equidistant from the lines whose angle they bisect. It can thus be seen that $P$ is equidistant from $AB, AD,$ and $CD$ and $Q$ is equidistant from $AB, BC,$ and $CD.$ If we let the feet of the altitudes from $P$ to $AB, AD,$ and $CD$ be called $E, F,$ and $G$ respectively, we can say that $PE = PF = PG.$ Analogously, we let the feet of the altitudes from $Q$ to $AB, BC,$ and $CD$ be $H, I,$ and $J$ respectively. Thus, $QH = QI = QJ.$ Because $ABCD$ is an isosceles trapezoid, we can say that all of the altitudes are equal to each other.

By SA as well as SS congruence for right triangles, we find that triangles $AEP, AFP, BHQ,$ and $BIQ$ are congruent. Similarly, $DFP, DGP, CJQ,$ and $CIQ$ by the same reasoning. Additionally, $EH = GJ = PQ$ since $EHQP$ and $GJQP$ are congruent rectangles.

If we then let $x = AE = AF = BH = BI,$ let $y = CI = CJ = DG = DF,$ and let $z = EH = GJ = PQ,$ we can create the following system of equations with the given side length information: $begin{align*} 2x + z &= 500, \ 2y + z &= 650, \ x + y &= 333. end{align*}$ Adding the first two equations, subtracting by twice the second, and dividing by $2$ yields $z = PQ = boxed{242}.$

~regular

Problem4

Let $w = dfrac{sqrt{3} + i}{2}$ and $z = dfrac{-1 + isqrt{3}}{2},$ where $i = sqrt{-1}.$ Find the number of ordered pairs $(r,s)$ of positive integers not exceeding $100$ that satisfy the equation $i cdot w^r = z^s.$

Solution 1

We rewrite $w$ and $z$ in polar form: $begin{align*} w &= e^{icdotfrac{pi}{6}}, \ z &= e^{icdotfrac{2pi}{3}}. end{align*}$ The equation $i cdot w^r = z^s$ becomes $begin{align*} e^{icdotfrac{pi}{2}} cdot left(e^{icdotfrac{pi}{6}}right)^r &= left(e^{icdotfrac{2pi}{3}}right)^s \ e^{ileft(frac{pi}{2}+frac{pi}{6}rright)} &= e^{ileft(frac{2pi}{3}sright)} \ frac{pi}{2}+frac{pi}{6}r &= frac{2pi}{3}s+2pi k \ 3+r &= 4s+12k \ 3+r &= 4(s+3k). end{align*}$ for some integer $k.$

Since $4leq 3+rleq 103$ and $4mid 3+r,$ we conclude that $begin{align*} 3+r &in {4,8,12,ldots,100}, \ s+3k &in {1,2,3,ldots,25}. end{align*}$ Note that the values for $s+3k$ and the values for $r$ have one-to-one correspondence.

We apply casework to the values for $s+3k:$

1. $s+3kequiv0pmod{3}$

There are $8$ values for $s+3k,$ so there are $8$ values for $r.$ It follows that $sequiv0pmod{3},$ so there are $33$ values for $s.$

There are $8cdot33=264$ ordered pairs $(r,s)$ in this case.

1. $s+3kequiv1pmod{3}$

There are $9$ values for $s+3k,$ so there are $9$ values for $r.$ It follows that $sequiv1pmod{3},$ so there are $34$ values for $s.$

There are $9cdot34=306$ ordered pairs $(r,s)$ in this case.

1. $s+3kequiv2pmod{3}$

There are $8$ values for $s+3k,$ so there are $8$ values for $r.$ It follows that $sequiv2pmod{3},$ so there are $33$ values for $s.$

There are $8cdot33=264$ ordered pairs $(r,s)$ in this case.

Together, the answer is $264+306+264=boxed{834}.$

~MRENTHUSIASM

Solution 2

First we recognize that $w = operatorname{cis}(30^{circ})$ and $z = operatorname{cis}(120^{circ})$ because the cosine and sine sums of those angles give the values of $w$ and $z$ , respectively. By Demoivre's theorem, $operatorname{cis}(theta)^n = operatorname{cis}(ntheta)$ . When you multiply by $i$ , we can think of that as rotating the complex number $90^{circ}$ counterclockwise in the complex plane. Therefore, by the equation we know that $30r + 90$ and $120s$ land on the same angle.

This means that $[30r + 90 equiv 120s pmod{360},]$ which we can simplify to $[r+3 equiv 4s pmod{12}.]$ Notice that this means that $r$ cycles by $12$ for every value of $s$ . This is because once $r$ hits $12$ , we get an angle of $360^{circ}$ and the angle laps onto itself again. By a similar reasoning, $s$ laps itself every $3$ times, which is much easier to count. By listing the possible values out, we get the pairs $(r,s)$ : $[begin{array}{cccccccc} (1,1) & (5,2) & (9,3) & (13,1) & (17,2) & (21,3) & ldots & (97,1) \ (1,4) & (5,5) & (9,6) & (13,4) & (17,5) & (21,6) & ldots & (97,4) \ (1,7) & (5,8) & (9,9) & (13,7) & (17,8) & (21,9) & ldots & (97,7) \ [-1ex] vdots & vdots & vdots & vdots & vdots & vdots & vdots & vdots \ (1,100) & (5,98) & (9,99) & (13,100) & (17,98) & (21,99) & ldots & (97,100) end{array}]$ We have $25$ columns in total: $34$ values for the first column, $33$ for the second, $33$ for the third, and then $34$ for the fourth, $33$ for the fifth, $33$ for the sixth, etc. Therefore, this cycle repeats every $3$ columns and our total sum is $(34+33+33) cdot 8 + 34 = 100 cdot 8 + 34 = boxed{834}$ .

~KingRavi

Problem5

A straight river that is $264$ meters wide flows from west to east at a rate of $14$ meters per minute. Melanie and Sherry sit on the south bank of the river with Melanie a distance of $D$ meters downstream from Sherry. Relative to the water, Melanie swims at $80$ meters per minute, and Sherry swims at $60$ meters per minute. At the same time, Melanie and Sherry begin swimming in straight lines to a point on the north bank of the river that is equidistant from their starting positions. The two women arrive at this point simultaneously. Find $D$ .

Solution 1 (Euclidean)

Define $m$ as the number of minutes they swim for.

Let their meeting point be $A$ . Melanie is swimming against the current, so she must aim upstream from point $A$ , to compensate for this; in particular, since she is swimming for $m$ minutes, the current will push her $14m$ meters downstream in that time, so she must aim for a point $B$ that is $14m$ meters upstream from point $A$ . Similarly, Sherry is swimming downstream for $m$ minutes, so she must also aim at point $B$ to compensate for the flow of the current.

If Melanie and Sherry were to both aim at point $B$ in a currentless river with the same dimensions, they would still both meet at that point simultaneously. Since there is no current in this scenario, the distances that Melanie and Sherry travel, respectively, are $80m$ and $60m$ meters. We can draw out this new scenario, with the dimensions that we have: $[asy] unitsize(0.02cm); draw((0,0)--(0,264)--(550,264)--(550,0)--cycle); pair B = (198,264); dot(B^^(0,0)^^(550,0),linewidth(5)); draw((0,0)--B,dashed); draw((550,0)--B,dashed); label("$60m$", (0,0)--B, E); label("$80m$", (550,0)--B, W); label("$264$", (0,0)--(0,264), W); label("$frac{D}{2} - 14m$", (0,264)--B, N); label("$frac{D}{2} + 14m$", B--(550,264), N); label("$D$", (0,0)--(550,0), S); [/asy]$ (While it is indeed true that the triangle above with side lengths $60m$ , $80m$ and $D$ is a right triangle, we do not know this yet, so we cannot assume this based on the diagram.)

By Pythagorean, we have $begin{align*} 264^{2} + left( frac{D}{2} - 14m right) ^{2} &= 3600m^{2} \ 264^{2} + left( frac{D}{2} + 14m right) ^{2} &= 6400m^{2}. end{align*}$

Subtracting the first equation from the second gives us $28Dm = 2800m^{2}$ , so $D = 100m$ . Substituting this into our first equation, we have that $begin{align*}264^{2} + 36^{2} m^{2} &= 60^{2}m^{2} \ 264^{2} &= 96 cdot 24 cdot m^{2} \ 11^{2} &= 4 cdot m^{2} \ m &= frac{11}{2}. end{align*}$

So $D = 100m = boxed{550}$ .

~ihatemath123

Solution 2 (Vectors)

We have the following diagram: $[asy] /* Made by MRENTHUSIASM */ size(350); pair A, B, C; A = (0,264); B = (-275,0); C = (275,0); draw((-300,0)--(300,0)^^(-300,264)--(300,264)^^A--B^^A--C,linewidth(2)); dot("Finish",A,1.75*N,linewidth(5)); dot("Sherry",B,1.75*S,linewidth(5)); dot("Melanie",C,1.75*S,linewidth(5)); Label L1 = Label("$D$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label("$264$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); Label L3 = Label("Current $(14)$", position=EndPoint, filltype=Fill(3,0,white)); Label L4 = Label("$y$", align=(-1,0), position=Relative(0.4)); Label L5 = Label("$x$", align=(0,1), position=Relative(0.4)); Label L6 = Label("$y$", align=(1,0), position=Relative(0.4)); Label L7 = Label("$x$", align=(0,1), position=Relative(0.4)); draw(B-(0,75)--C-(0,75), L=L1, arrow=Arrows(),bar=Bars(15)); draw((-350,0)--(-350,264), L=L2, arrow=Arrows(),bar=Bars(15)); draw((-300,-120)--(300,-120), L=L3, arrow=EndArrow()); draw(B--B+(0,48), L=L4, arrow=EndArrow()); draw(B+(0,48)--B+(50,48), L=L5, arrow=EndArrow()); draw(C--C+(0,48), L=L6, arrow=EndArrow()); draw(C+(0,48)--C+(-50,48), L=L7, arrow=EndArrow()); [/asy]$ Since Melanie and Sherry swim for the same distance and the same amount of time, they swim at the same net speed.

Let $x$ and $y$ be some positive numbers. We have the following table: $[begin{array}{c||c|c|c} & textbf{Net Velocity Vector (m/min)} & textbf{Natural Velocity Vector (m/min)} & textbf{Natural Speed (m/min)} \ hline hline &&& \ [-2.25ex] textbf{Melanie} & langle -x,yrangle & langle -x-14,yrangle & 80 \ hline &&& \ [-2.25ex] textbf{Sherry} & langle x,yrangle & langle x-14,yrangle & 60 end{array}]$ Recall that $|text{velocity}|=text{speed},$ so $begin{align*} (-x-14)^2 + y^2 &= 80^2, &&(1) \ (x-14)^2 + y^2 &= 60^2. &&(2) end{align*}$ We subtract $(2)$ from $(1)$ to get $56x=2800,$ from which $x=50.$ Substituting this into either equation, we have $y=48.$

It follows that Melanie and Sherry both swim for $264div y=5.5$ minutes. Therefore, the answer is $[D=2xcdot5.5=boxed{550}.]$ ~MRENTHUSIASM

Problem5

Find the number of ordered pairs of integers $(a, b)$ such that the sequence $[3, 4, 5, a, b, 30, 40, 50]$ is strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression.

Solution 1

Since $3,4,5,a$ and $3,4,5,b$ cannot be an arithmetic progression, $a$ or $b$ can never be $6$ . Since $b, 30, 40, 50$ and $a, 30, 40, 50$ cannot be an arithmetic progression, $b$ can never be $20$ . Since $a < b$ , there are ${24 - 2 choose 2} = 231$ ways to choose $a$ and $b$ with these two restrictions in mind.

However, there are still specific invalid cases counted in these $231$ pairs $(a,b)$ . Since $[3,5,a,b]$ cannot form an arithmetic progression, $underline{(a,b) neq (7,9)}$ . $[a,b,30,50]$ cannot be an arithmetic progression, so $(a,b) neq (-10,10)$ ; however, since this pair was not counted in our $231$ , we do not need to subtract it off. $[3,a,b,30]$ cannot form an arithmetic progression, so $underline{(a,b) neq (12,21)}$ . $[4, a, b, 40]$ cannot form an arithmetic progression, so $underline{(a,b) neq (16,28)}$ . $[5, a,b, 50]$ cannot form an arithmetic progression, $(a,b) neq 20, 35$ ; however, since this pair was not counted in our $231$ (since we disallowed $a$ or $b$ to be $20$ ), we do not to subtract it off.

Also, the sequences $(3,a,b,40)$ , $(3,a,b,50)$ , $(4,a,b,30)$ , $(4,a,b,50)$ , $(5,a,b,30)$ and $(5,a,b,40)$ will never be arithmetic, since that would require $a$ and $b$ to be non-integers.

So, we need to subtract off $3$ progressions from the $231$ we counted, to get our final answer of $boxed{228}$ .

~ ihatemath123

Solution 2 (Rigorous)

We will follow the solution from earlier in a rigorous manner to show that there are no other cases missing.

We recognize that an illegal sequence (defined as one that we subtract from our 231) can never have the numbers {3, 4} and {4,5} because we have not included a 6 in our count. Similarly, sequences with {30,40} and {40,50} will not give us any subtractions because those sequences must all include a 20. Let's stick with the lower ones for a minute: if we take them two at a time, then {3,5} will give us the subtraction of 1 sequence {3,5,7,9}. We have exhausted all pairs of numbers we can take, and if we take the triplet of single digit numbers, the only possible sequence must have a 6, which we already don't count. Therefore, we subtract $textbf{1}$ from the count of illegal sequences with any of the single-digit numbers and none of the numbers 30,40,50. (Note if we take only 1 at a time, there will have to be 3 of $a, b$ , which is impossible.)

If we have the sequence including {30,50}, we end up having negative values, so these do not give us any subtractions, and the triplet {30,40,50} gives us a 20. Hence by the same reasoning as earlier, we have 0 subtractions from the sequences with these numbers and none of the single digit numbers {3,4,5}.

Finally, we count the sequences that are something like (one of 3,4,5,), $a, b$ , (one of 30, 40, 50). If this is to be the case, then let $a$ be the starting value in the sequence. The sequence will be $a, a+d, a+2d, a+3d$ ; We see that if we subtract the largest term by the smallest term we have $3d$ , so the subtraction of one of (30,40,50) and one of (3,4,5) must be divisible by 3. Therefore the only sequences possible are $3,a,b,30; 4,a,b,40; 5,a,b,50$ . Of these, only the last is invalid because it gives $b = 35$ , larger than our bounds $6<a<b<30$ . Therefore, we subtract $textbf{2}$ from this case.

Our final answer is $231 - 1 - 2 = boxed{228}$

~KingRavi

Solution 3

Denote $S = left{ (a, b) : 6 leq a < b leq 29 right}$ .

Denote by $A$ a subset of $S$ , such that there exists an arithmetic sequence that has 4 terms and includes $a$ but not $b$ .

Denote by $B$ a subset of $S$ , such that there exists an arithmetic sequence that has 4 terms and includes $b$ but not $a$ .

Hence, $C$ is a subset of $S$ , such that there exists an arithmetic sequence that has 4 terms and includes both $a$ and $b$ .

Hence, this problem asks us to compute $[ | S | - left( | A | + | B | + | C | right) . ]$

First, we compute $| S |$ .

We have $| S | = binom{29 - 6 + 1}{2} = binom{24}{2} = 276$ .

Second, we compute $| A |$ .

$textbf{Case 1}$ : $a = 6$ .

We have $b = 8 , cdots , 19, 21, 22, cdots, 29$ . Thus, the number of solutions is 21.

$textbf{Case 2}$ : $a = 20$ .

We have $b = 21, 22, cdots , 29$ . Thus, the number of solutions is 9.

Thus, $| A | = 21 + 9 = 30$ .

Third, we compute $| B |$ .

In $B$ , we have $b = 6, 20$ . However, because $6 leq a < b$ , we have $b geq 7$ . Thus, $b = 20$ .

This implies $a = 7, 8, 9, 11, 12, cdots , 19$ . Thus, $| B | = 12$ .

Fourth, we compute $| C |$ .

$textbf{Case 1}$ : In the arithmetic sequence, the two numbers beyond $a$ and $b$ are on the same side of $a$ and $b$ .

Hence, $(a, b) = (6 , 7), (7, 9) , (10, 20)$ . Therefore, the number solutions in this case is 3.

$textbf{Case 2}$ : In the arithmetic sequence, the two numbers beyond $a$ and $b$ are on the opposite sides of $a$ and $b$ .

$textbf{Case 2.1}$ : The arithmetic sequence is $3, a, b, 30$ .

Hence, $(a, b) = (12, 21)$ .

$textbf{Case 2.2}$ : The arithmetic sequence is $4, a, b, 40$ .

Hence, $(a, b) = (16, 28)$ .

$textbf{Case 2.3}$ : The arithmetic sequence is $5, a, b, 50$ .

Hence, $(a, b) = (20, 35)$ .

Putting two cases together, $| C | = 6$ .

Therefore, $[| S | - left( | A | + | B | + | C | right) = 276 - left( 30 + 12 + 6 right) = boxed{228}.]$

~Steven Chen (www.professorchenedu.com)

Solution 4

divide cases into $7leq a<20; 21leq aleq28$ .(Notice that $a$ can't be equal to $6,20$ , that's why I divide them into two parts. There are three cases that arithmetic sequence forms: $3,12,21,30;4,16,28,40;3,5,7,9$ .(NOTICE that $5,20,35,50$ IS NOT A VALID SEQUENCE!) So when $7leq a<20$ , there are $10+11+12+...+22-3-13=192$ possible ways( 3 means the arithmetic sequence and 13 means there are 13 "a" s and b cannot be 20)

When $21leq a leq 28$ , there are $1+2+cdots+8=36$ ways.

In all, there are $192+36=boxed{228}$ possible sequences.

~bluesoul

Problem6

Solution 1

Also, the sequences $(3,a,b,40)$ , $(3,a,b,50)$ , $(4,a,b,30)$ , $(4,a,b,50)$ , $(5,a,b,30)$ and $(5,a,b,40)$ will never be arithmetic, since that would require $a$ and $b$ to be non-integers.

So, we need to subtract off $3$ progressions from the $231$ we counted, to get our final answer of $boxed{228}$ .

~ ihatemath123

Solution 2 (Rigorous)

We will follow the solution from earlier in a rigorous manner to show that there are no other cases missing.

Our final answer is $231 - 1 - 2 = boxed{228}$

~KingRavi

Solution 3

Denote $S = left{ (a, b) : 6 leq a < b leq 29 right}$ .

Denote by $A$ a subset of $S$ , such that there exists an arithmetic sequence that has 4 terms and includes $a$ but not $b$ .

Denote by $B$ a subset of $S$ , such that there exists an arithmetic sequence that has 4 terms and includes $b$ but not $a$ .

Hence, $C$ is a subset of $S$ , such that there exists an arithmetic sequence that has 4 terms and includes both $a$ and $b$ .

Hence, this problem asks us to compute $[ | S | - left( | A | + | B | + | C | right) . ]$

First, we compute $| S |$ .

We have $| S | = binom{29 - 6 + 1}{2} = binom{24}{2} = 276$ .

Second, we compute $| A |$ .

$textbf{Case 1}$ : $a = 6$ .

We have $b = 8 , cdots , 19, 21, 22, cdots, 29$ . Thus, the number of solutions is 21.

$textbf{Case 2}$ : $a = 20$ .

We have $b = 21, 22, cdots , 29$ . Thus, the number of solutions is 9.

Thus, $| A | = 21 + 9 = 30$ .

Third, we compute $| B |$ .

In $B$ , we have $b = 6, 20$ . However, because $6 leq a < b$ , we have $b geq 7$ . Thus, $b = 20$ .

This implies $a = 7, 8, 9, 11, 12, cdots , 19$ . Thus, $| B | = 12$ .

Fourth, we compute $| C |$ .

$textbf{Case 1}$ : In the arithmetic sequence, the two numbers beyond $a$ and $b$ are on the same side of $a$ and $b$ .

Hence, $(a, b) = (6 , 7), (7, 9) , (10, 20)$ . Therefore, the number solutions in this case is 3.

$textbf{Case 2}$ : In the arithmetic sequence, the two numbers beyond $a$ and $b$ are on the opposite sides of $a$ and $b$ .

$textbf{Case 2.1}$ : The arithmetic sequence is $3, a, b, 30$ .

Hence, $(a, b) = (12, 21)$ .

$textbf{Case 2.2}$ : The arithmetic sequence is $4, a, b, 40$ .

Hence, $(a, b) = (16, 28)$ .

$textbf{Case 2.3}$ : The arithmetic sequence is $5, a, b, 50$ .

Hence, $(a, b) = (20, 35)$ .

Putting two cases together, $| C | = 6$ .

Therefore, $[| S | - left( | A | + | B | + | C | right) = 276 - left( 30 + 12 + 6 right) = boxed{228}.]$

~Steven Chen (www.professorchenedu.com)

Solution 4

When $21leq a leq 28$ , there are $1+2+cdots+8=36$ ways.

In all, there are $192+36=boxed{228}$ possible sequences.

~bluesoul

Problem7

Let $a,b,c,d,e,f,g,h,i$ be distinct integers from $1$ to $9.$ The minimum possible positive value of $[dfrac{a cdot b cdot c - d cdot e cdot f}{g cdot h cdot i}]$ can be written as $frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution

To minimize a positive fraction, we minimize its numerator and maximize its denominator. It is clear that $frac{a cdot b cdot c - d cdot e cdot f}{g cdot h cdot i} geq frac{1}{7cdot8cdot9}.$

If we minimize the numerator, then $a cdot b cdot c - d cdot e cdot f = 1.$ Note that $a cdot b cdot c cdot d cdot e cdot f = (a cdot b cdot c) cdot (a cdot b cdot c - 1) geq 6! = 720,$ so $a cdot b cdot c geq 28.$ It follows that $a cdot b cdot c$ and $d cdot e cdot f$ are consecutive composites with prime factors no other than $2,3,5,$ and $7.$ The smallest values for $a cdot b cdot c$ and $d cdot e cdot f$ are $36$ and $35,$ respectively. So, we have ${a,b,c} = {2,3,6}, {d,e,f} = {1,5,7},$ and ${g,h,i} = {4,8,9},$ from which $frac{a cdot b cdot c - d cdot e cdot f}{g cdot h cdot i} = frac{1}{288}.$

If we do not minimize the numerator, then $a cdot b cdot c - d cdot e cdot f > 1.$ Note that $frac{a cdot b cdot c - d cdot e cdot f}{g cdot h cdot i} geq frac{2}{7cdot8cdot9} > frac{1}{288}.$

Together, we conclude that the minimum possible positive value of $frac{a cdot b cdot c - d cdot e cdot f}{g cdot h cdot i}$ is $frac{1}{288}.$ Therefore, the answer is $1+288=boxed{289}.$

~MRENTHUSIASM ~jgplay

Problem8

Equilateral triangle $triangle ABC$ is inscribed in circle $omega$ with radius $18.$ Circle $omega_A$ is tangent to sides $overline{AB}$ and $overline{AC}$ and is internally tangent to $omega.$ Circles $omega_B$ and $omega_C$ are defined analogously. Circles $omega_A,$ $omega_B,$ and $omega_C$ meet in six points---two points for each pair of circles. The three intersection points closest to the vertices of $triangle ABC$ are the vertices of a large equilateral triangle in the interior of $triangle ABC,$ and the other three intersection points are the vertices of a smaller equilateral triangle in the interior of $triangle ABC.$ The side length of the smaller equilateral triangle can be written as $sqrt{a} - sqrt{b},$ where $a$ and $b$ are positive integers. Find $a+b.$

Diagram

$[asy] /* Made by MRENTHUSIASM */ size(250); pair A, B, C, W, WA, WB, WC, X, Y, Z; A = 18*dir(90); B = 18*dir(210); C = 18*dir(330); W = (0,0); WA = 6*dir(270); WB = 6*dir(30); WC = 6*dir(150); X = (sqrt(117)-3)*dir(270); Y = (sqrt(117)-3)*dir(30); Z = (sqrt(117)-3)*dir(150); filldraw(X--Y--Z--cycle,green,dashed); draw(Circle(WA,12)^^Circle(WB,12)^^Circle(WC,12),blue); draw(Circle(W,18)^^A--B--C--cycle); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$omega$",W,1.5*dir(270),linewidth(4)); dot("$omega_A$",WA,1.5*dir(-WA),linewidth(4)); dot("$omega_B$",WB,1.5*dir(-WB),linewidth(4)); dot("$omega_C$",WC,1.5*dir(-WC),linewidth(4)); [/asy]$ ~MRENTHUSIASM ~ihatemath123

Solution 1 (Coordinate Geometry)

We can extend $AB$ and $AC$ to $B'$ and $C'$ respectively such that circle $omega_A$ is the incircle of $triangle AB'C'$ . $[asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, B1, C1, W, WA, WB, WC, X, Y, Z; A = 18*dir(90); B = 18*dir(210); C = 18*dir(330); B1 = A+24*sqrt(3)*dir(B-A); C1 = A+24*sqrt(3)*dir(C-A); W = (0,0); WA = 6*dir(270); WB = 6*dir(30); WC = 6*dir(150); X = (sqrt(117)-3)*dir(270); Y = (sqrt(117)-3)*dir(30); Z = (sqrt(117)-3)*dir(150); filldraw(X--Y--Z--cycle,green,dashed); draw(Circle(WA,12)^^Circle(WB,12)^^Circle(WC,12),blue); draw(Circle(W,18)^^A--B--C--cycle); draw(B--B1--C1--C,dashed); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*(-1,0),linewidth(4)); dot("$C$",C,1.5*(1,0),linewidth(4)); dot("$B'$",B1,1.5*dir(B1),linewidth(4)); dot("$C'$",C1,1.5*dir(C1),linewidth(4)); dot("$O$",W,1.5*dir(90),linewidth(4)); dot("$X$",X,1.5*dir(X),linewidth(4)); dot("$Y$",Y,1.5*dir(Y),linewidth(4)); dot("$Z$",Z,1.5*dir(Z),linewidth(4)); [/asy]$ Since the diameter of the circle is the height of this triangle, the height of this triangle is $36$ . We can use inradius or equilateral triangle properties to get the inradius of this triangle is $12$ (The incenter is also a centroid in an equilateral triangle, and the distance from a side to the centroid is a third of the height). Therefore, the radius of each of the smaller circles is $12$ .

Let $O=omega$ be the center of the largest circle. We will set up a coordinate system with $O$ as the origin. The center of $omega_A$ will be at $(0,-6)$ because it is directly beneath $O$ and is the length of the larger radius minus the smaller radius, or $18-12 = 6$ . By rotating this point $120^{circ}$ around $O$ , we get the center of $omega_B$ . This means that the magnitude of vector $overrightarrow{Oomega_B}$ is $6$ and is at a $30$ degree angle from the horizontal. Therefore, the coordinates of this point are $(3sqrt{3},3)$ and by symmetry the coordinates of the center of $omega_C$ is $(-3sqrt{3},3)$ .

The upper left and right circles intersect at two points, the lower of which is $X$ . The equations of these two circles are: $begin{align*} (x+3sqrt3)^2 + (y-3)^2 &= 12^2, \ (x-3sqrt3)^2 + (y-3)^2 &= 12^2. end{align*}$ We solve this system by subtracting to get $x = 0$ . Plugging back in to the first equation, we have $(3sqrt{3})^2 + (y-3)^2 = 144 implies (y-3)^2 = 117 implies y-3 = pm sqrt{117} implies y = 3 pm sqrt{117}$ . Since we know $X$ is the lower solution, we take the negative value to get $X = (0,3-sqrt{117})$ .

We can solve the problem two ways from here. We can find $Y$ by rotation and use the distance formula to find the length, or we can be somewhat more clever. We notice that it is easier to find $OX$ as they lie on the same vertical, $angle XOY$ is $120$ degrees so we can make use of $30-60-90$ triangles, and $OX = OY$ because $O$ is the center of triangle $XYZ$ . We can draw the diagram as such: $[asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, B1, C1, W, WA, WB, WC, X, Y, Z; A = 18*dir(90); B = 18*dir(210); C = 18*dir(330); B1 = A+24*sqrt(3)*dir(B-A); C1 = A+24*sqrt(3)*dir(C-A); W = (0,0); WA = 6*dir(270); WB = 6*dir(30); WC = 6*dir(150); X = (sqrt(117)-3)*dir(270); Y = (sqrt(117)-3)*dir(30); Z = (sqrt(117)-3)*dir(150); filldraw(X--Y--Z--cycle,green,dashed); draw(Circle(WA,12)^^Circle(WB,12)^^Circle(WC,12),blue); draw(Circle(W,18)^^A--B--C--cycle); draw(B--B1--C1--C^^W--X^^W--Y^^W--midpoint(X--Y),dashed); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*(-1,0),linewidth(4)); dot("$C$",C,1.5*(1,0),linewidth(4)); dot("$B'$",B1,1.5*dir(B1),linewidth(4)); dot("$C'$",C1,1.5*dir(C1),linewidth(4)); dot("$O$",W,1.5*dir(90),linewidth(4)); dot("$X$",X,1.5*dir(X),linewidth(4)); dot("$Y$",Y,1.5*dir(Y),linewidth(4)); dot("$Z$",Z,1.5*dir(Z),linewidth(4)); [/asy]$ Note that $OX = OY = sqrt{117} - 3$ . It follows that $begin{align*} XY &= 2 cdot frac{OXcdotsqrt{3}}{2} \ &= OX cdot sqrt{3} \ &= (sqrt{117}-3) cdot sqrt{3} \ &= sqrt{351}-sqrt{27}. end{align*}$ Finally, the answer is $351+27 = boxed{378}$ .

~KingRavi

Solution 2 (Euclidean Geometry)

$[asy] /* Made by MRENTHUSIASM */ /* Modified by isabelchen */ size(250); pair A, B, C, W, WA, WB, WC, X, Y, Z, D, E; A = 18*dir(90); B = 18*dir(210); C = 18*dir(330); W = (0,0); WA = 6*dir(270); WB = 6*dir(30); WC = 6*dir(150); X = (sqrt(117)-3)*dir(270); Y = (sqrt(117)-3)*dir(30); Z = (sqrt(117)-3)*dir(150); D = intersectionpoint(Circle(WA,12),A--C); E = intersectionpoints(Circle(WB,12),Circle(WC,12))[0]; filldraw(X--Y--Z--cycle,green,dashed); draw(Circle(WA,12)^^Circle(WB,12)^^Circle(WC,12),blue); draw(Circle(W,18)^^A--B--C--cycle); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$omega$",W,1.5*dir(270),linewidth(4)); dot("$omega_A$",WA,1.5*dir(-WA),linewidth(4)); dot("$omega_B$",WB,1.5*dir(-WB),linewidth(4)); dot("$omega_C$",WC,1.5*dir(-WC),linewidth(4)); dot("$X$",X,1.5*dir(X),linewidth(4)); dot("$Y$",Y,1.5*dir(Y),linewidth(4)); dot("$Z$",Z,1.5*dir(Z),linewidth(4)); dot("$E$",E,1.5*dir(E),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); draw(WC--WB^^WC--X^^WC--E^^WA--D^^A--X); [/asy]$ For equilateral triangle with side length $l$ , height $h$ , and circumradius $r$ , there are relationships: $h = frac{sqrt{3}}{2} l$ , $r = frac{2}{3} h = frac{sqrt{3}}{3} l$ , and $l = sqrt{3}r$ .

There is a lot of symmetry in the figure. The radius of the big circle $odot omega$ is $R = 18$ , let the radius of the small circles $odot omega_A$ , $odot omega_B$ , $odot omega_C$ be $r$ .

We are going to solve this problem in $3$ steps:

$textbf{Step 1:}$

We have $triangle A omega_A D$ is a $30-60-90$ triangle, and $A omega_A = 2 cdot omega_A D$ , $A omega_A = 2R-r$ ( $odot omega$ and $odot omega_A$ are tangent), and $omega_A D = r$ . So, we get $2R-r = 2r$ and $r = frac{2}{3} cdot R = 12$ .

Since $odot omega$ and $odot omega_A$ are tangent, we get $omega omega_A = R - r = frac{1}{3} cdot R = 6$ .

Note that $triangle omega_A omega_B omega_C$ is an equilateral triangle, and $omega$ is its center, so $omega_B omega_C = sqrt{3} cdot omega omega_A = 6 sqrt{3}$ .

$textbf{Step 2:}$

Note that $triangle omega_C E X$ is an isosceles triangle, so $[EX = 2 sqrt{(omega_C E)^2 - left(frac{omega_B omega_C}{2}right)^2} = 2 sqrt{r^2 - left(frac{omega_B omega_C}{2}right)^2} = 2 sqrt{12^2 - (3 sqrt{3})^2} = 2 sqrt{117}.]$

$textbf{Step 3:}$

In $odot omega_C$ , Power of a Point gives $omega X cdot omega E = r^2 - (omega_C omega)^2$ and $omega E = EX - omega X = 2sqrt{117} - omega X$ .

It follows that $omega X cdot (2sqrt{117} - omega X) = 12^2 - 6^2$ . We solve this quadratic equation: $omega X = sqrt{117} - 3$ .

Since $omega X$ is the circumradius of equilateral $triangle XYZ$ , we have $XY = sqrt{3} cdot omega X = sqrt{3} cdot (sqrt{117} - 3) = sqrt{351}-sqrt{27}$ .

Therefore, the answer is $351+27 = boxed{378}$ .

~isabelchen

Problem9

Ellina has twelve blocks, two each of red ( $textbf{R}$ ), blue ( $textbf{B}$ ), yellow ( $textbf{Y}$ ), green ( $textbf{G}$ ), orange ( $textbf{O}$ ), and purple ( $textbf{P}$ ). Call an arrangement of blocks $textit{even}$ if there is an even number of blocks between each pair of blocks of the same color. For example, the arrangement $[textbf{R B B Y G G Y R O P P O}]$ is even. Ellina arranges her blocks in a row in random order. The probability that her arrangement is even is $frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution 1

Consider this position chart: $[textbf{1 2 3 4 5 6 7 8 9 10 11 12}]$ Since there has to be an even number of spaces between each ball of the same color, spots $1$ , $3$ , $5$ , $7$ , $9$ , and $11$ contain some permutation of all 6 colored balls. Likewise, so do the even spots, so the number of even configurations is $6! cdot 6!$ (after putting every pair of colored balls in opposite parity positions, the configuration can be shown to be even). This is out of $frac{12!}{(2!)^6}$ possible arrangements, so the probability is: $[frac{6!cdot6!}{frac{12!}{(2!)^6}} = frac{6!cdot2^6}{7cdot8cdot9cdot10cdot11cdot12} = frac{2^4}{7cdot11cdot3} = frac{16}{231},]$ which is in simplest form. So, $m + n = 16 + 231 = boxed{247}$ .

~Oxymoronic15

Solution 2

We can simply use constructive counting. First, let us place the red balls; choose the first slot in $12$ ways, and the second in $6$ ways, because the number is cut in half due to the condition in the problem. This gives $12 cdot 6$ ways to place the blue balls. Similarly, there are $10 cdot 5$ ways to place the blue balls, and so on, until there are $2 cdot 1$ ways to place the purple balls. Thus, the probability is $[frac{12 cdot 6 cdot 10 cdot 5 cdot 8 cdot 4 cdot 6 cdot 3 cdot 4 cdot 2 cdot 2 cdot 1}{12!}=frac{16}{231},]$ and the desired answer extraction is $16+231=boxed{textbf{247}}$ .

~A1001

Problem10

Three spheres with radii $11$ , $13$ , and $19$ are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at $A$ , $B$ , and $C$ , respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that $AB^2 = 560$ . Find $AC^2$ .

Diagrams

$[asy] size(500); pair A, B, OA, OB; B = (0,0); A = (-23.6643191,0); OB = (0,-8); OA = (-23.6643191,-4); draw(circle(OB,13)); draw(circle(OA,11)); draw((-48,0)--(24,0)); label("$l$",(-42,1),N); label("$A$",A,N); label("$B$",B,N); label("$O_A$",OA,S); label("$O_B$",OB,S); draw(A--OA); draw(B--OB); draw(OA--OB); draw(OA--(0,-4)); draw(OA--(-33.9112699,0)); draw(OB--(10.2469508,0)); label("$24$",midpoint(OA--OB),S); label("$sqrt{560}$",midpoint(A--B),N); label("$11$",midpoint(OA--(-33.9112699,0)),S); label("$13$",midpoint(OB--(10.2469508,0)),S); label("$r$",midpoint(midpoint(A--B)--A),N); label("$r$",midpoint(midpoint(A--B)--B),N); label("$r$",midpoint(A--(-33.9112699,0)),N); label("$r$",midpoint(B--(10.2469508,0)),N); label("$x$",midpoint(midpoint(B--OB)--OB),E); label("$D$",midpoint(B--OB),E); [/asy]$

$[asy] size(500); pair A, C, OA, OC; C = (0,0); A = (-27.4954541697,0); OC = (0,-16); OA = (-27.4954541697,-4); draw(circle(OC,19)); draw(circle(OA,11)); draw((-48,0)--(24,0)); label("$l$",(-42,1),N); label("$A$",A,N); label("$C$",C,N); label("$O_A$",OA,S); label("$O_C$",OC,S); draw(A--OA); draw(C--OC); draw(OA--OC); draw(OA--(0,-4)); draw(OA--(-37.8877590151,0)); draw(OC--(10.2469508,0)); label("$30$",midpoint(OA--OC),S); label("$11$",midpoint(OA--(-37.8877590151,0)),S); label("$19$",midpoint(OC--(10.2469508,0)),E); label("$r$",midpoint(midpoint(A--C)--A),N); label("$r$",midpoint(midpoint(A--C)--C),N); label("$r$",midpoint(A--(-37.8877590151,0)),N); label("$r$",midpoint(C--(10.2469508,0)),N); label("$E$",(0,-4),E); [/asy]$

Solution 1

We let $l$ be the plane that passes through the spheres and $O_A$ and $O_B$ be the centers of the spheres with radii $11$ and $13$ . We take a cross-section that contains $A$ and $B$ , which contains these two spheres but not the third. Because the plane cuts out congruent circles, they have the same radius and from the given information, $AB = sqrt{560}$ . Since $ABO_BO_A$ is a trapezoid, we can drop an altitude from $O_A$ to $BO_B$ to create a rectangle and triangle to use Pythagorean theorem. We know that the length of the altitude is $sqrt{560}$ and let the distance from $O_B$ to $D$ be $x$ . Then we have $x^2 = 576-560 implies x = 4$ .

We have $AO_A = BD$ because of the rectangle, so $sqrt{11^2-r^2} = sqrt{13^2-r^2}-4$ . Squaring, we have $121-r^2 = 169-r^2 + 16 - 8 cdot sqrt{169-r^2}$ . Subtracting, we get $8 cdot sqrt{169-r^2} = 64 implies sqrt{169-r^2} = 8 implies 169-r^2 = 64 implies r^2 = 105$ . We also notice that since we had $sqrt{169-r^2} = 8$ means that $BO_B = 8$ and since we know that $x = 4$ , $AO_A = 4$ .

We now look at our second diagram.

$CO_C = sqrt{19^2-r^2} = sqrt{361 - 105} = sqrt{256} = 16$ . Since $AO_A = 4$ , we have $EO_C = 16-4 = 12$ . Using Pythagorean theorem, $O_AE = sqrt{30^2 - 12^2} = sqrt{900-144} = sqrt{756}$ . Therefore, $O_AE^2 = AC^2 = boxed{756}$

~KingRavi

Solution 2

Let the distance between the center of the sphere to the center of those circular intersections as $a,b,c$ separately. $a-11,b-13,c-19$ . According to the problem, we have $a^2-11^2=b^2-13^2=c^2-19^2;(11+13)^2-(b-a)^2=560$ . After solving we have $b-a=4$ , plug this back to $11^2-a^2=13^2-b^2; a=4,b=8,c=16$

The desired value is $(11+19)^2-(16-4)^2=boxed{756}$

~bluesoul

Solution 3

Denote by $r$ the radius of three congruent circles formed by the cutting plane. Denote by $O_A$ , $O_B$ , $O_C$ the centers of three spheres that intersect the plane to get circles centered at $A$ , $B$ , $C$ , respectively.

Because three spheres are mutually tangent, $O_A O_B = 11 + 13 = 24$ , $O_A O_C = 11 + 19 = 32$ .

We have $O_A A^2 = 11^2 - r^2$ , $O_B B^2 = 13^2 - r^2$ , $O_C C^2 = 19^2 - r^2$ .

Because $O_A A$ and $O_B B$ are perpendicular to the plane, $O_A AB O_B$ is a right trapezoid, with $angle O_A A B = angle O_B BA = 90^circ$ .

Hence, $begin{align*} O_B B - O_A A & = sqrt{O_A O_B^2 - AB^2} \ & = 4 . hspace{1cm} (1) end{align*}$

Recall that $begin{align*} O_B B^2 - O_A A^2 & = left( 13^2 - r^2 right) - left( 11^2 - r^2 right) \ & = 48 . hspace{1cm} (2) end{align*}$

Hence, taking $frac{(2)}{(1)}$ , we get $[ O_B B + O_A A = 12 . hspace{1cm} (3) ]$

Solving (1) and (3), we get $O_B B = 8$ and $O_A A = 4$ .

Thus, $r^2 = 11^2 - O_A A^2 = 105$ .

Thus, $O_C C = sqrt{19^2 - r^2} = 16$ .

Because $O_A A$ and $O_C C$ are perpendicular to the plane, $O_A AC O_C$ is a right trapezoid, with $angle O_A A C = angle O_C CA = 90^circ$ .

Therefore, $begin{align*} AC^2 & = O_A O_C^2 - left( O_C C - O_A A right)^2 \ & = boxed{textbf{(756) }} . end{align*}$

$textbf{FINAL NOTE:}$ In our solution, we do not use the conditio that spheres $A$ and $B$ are externally tangent. This condition is redundant in solving this problem.

~Steven Chen (www.professorcheneeu.com)

Problem10

Diagrams

Solution 1

We now look at our second diagram.

~KingRavi

Solution 2

The desired value is $(11+19)^2-(16-4)^2=boxed{756}$

~bluesoul

Solution 3

Because three spheres are mutually tangent, $O_A O_B = 11 + 13 = 24$ , $O_A O_C = 11 + 19 = 32$ .

We have $O_A A^2 = 11^2 - r^2$ , $O_B B^2 = 13^2 - r^2$ , $O_C C^2 = 19^2 - r^2$ .

Because $O_A A$ and $O_B B$ are perpendicular to the plane, $O_A AB O_B$ is a right trapezoid, with $angle O_A A B = angle O_B BA = 90^circ$ .

Hence, $begin{align*} O_B B - O_A A & = sqrt{O_A O_B^2 - AB^2} \ & = 4 . hspace{1cm} (1) end{align*}$

Recall that $begin{align*} O_B B^2 - O_A A^2 & = left( 13^2 - r^2 right) - left( 11^2 - r^2 right) \ & = 48 . hspace{1cm} (2) end{align*}$

Hence, taking $frac{(2)}{(1)}$ , we get $[ O_B B + O_A A = 12 . hspace{1cm} (3) ]$

Solving (1) and (3), we get $O_B B = 8$ and $O_A A = 4$ .

Thus, $r^2 = 11^2 - O_A A^2 = 105$ .

Thus, $O_C C = sqrt{19^2 - r^2} = 16$ .

Because $O_A A$ and $O_C C$ are perpendicular to the plane, $O_A AC O_C$ is a right trapezoid, with $angle O_A A C = angle O_C CA = 90^circ$ .

Therefore, $begin{align*} AC^2 & = O_A O_C^2 - left( O_C C - O_A A right)^2 \ & = boxed{textbf{(756) }} . end{align*}$

$textbf{FINAL NOTE:}$ In our solution, we do not use the conditio that spheres $A$ and $B$ are externally tangent. This condition is redundant in solving this problem.

Problem11

Let $ABCD$ be a parallelogram with $angle BAD < 90^{circ}$ . A circle tangent to sides $overline{DA}$ , $overline{AB}$ , and $overline{BC}$ intersects diagonal $overline{AC}$ at points $P$ and $Q$ with $AP < AQ$ , as shown. Suppose that $AP = 3$ , $PQ = 9$ , and $QC = 16$ . Then the area of $ABCD$ can be expressed in the form $msqrt n$ , where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$ .

$[asy] defaultpen(linewidth(0.6)+fontsize(11)); size(8cm); pair A,B,C,D,P,Q; A=(0,0); label("$A$", A, SW); B=(6,15); label("$B$", B, NW); C=(30,15); label("$C$", C, NE); D=(24,0); label("$D$", D, SE); P=(5.2,2.6); label("$P$", (5.8,2.6), N); Q=(18.3,9.1); label("$Q$", (18.1,9.7), W); draw(A--B--C--D--cycle); draw(C--A); draw(Circle((10.95,7.45), 7.45)); dot(A^^B^^C^^D^^P^^Q); [/asy]$

Solution 1 (No trig)

Let's redraw the diagram, but extend some helpful lines.

$[asy] size(20cm); pair A,B,C,D,E,F,P,Q,O; A=(0,0); E = (24,15); F = (30,0); O = (10.5,7.5); label("$A$", A, SW); B=(6,15); label("$B$", B, NW); C=(30,15); label("$C$", C, NE); D=(24,0); label("$D$", D, SE); P=(5.2,2.6); label("$P$", (5.8,2.6), N); Q=(18.3,9.1); label("$Q$", (18.1,9.7), W); draw(A--B--C--D--cycle); draw(C--A); draw(Circle((10.95,7.45), 7.45)); dot(A^^B^^C^^D^^P^^Q); dot(O); label("$O$",O,W); draw((10.5,15)--(10.5,0)); draw(D--(24,15),dashed); draw(C--(30,0),dashed); draw(D--(30,0)); dot(E); dot(F); label("$3$", midpoint(A--P), S); label("$9$", midpoint(P--Q), S); label("$16$", midpoint(Q--C), S); label("$x$", (5.5,13.75), W); label("$20$", (20.25,15), N); label("$6$", (5.25,0), S); label("$6$", (1.5,3.75), W); label("$x$", (8.25,15),N); label("$14+x$", (17.25,0), S); label("$6-x$", (27,15), N); label("$6+x$", (27,7.5), W); label("$6sqrt{3}$", (30,7.5),W); label("$T_1$", (10.5,15), N); label("$T_2$", (10.5,0), S); label("$T_3$", (4.5,11.25),W); label("$E$",E, N); label("$F$",F, S); [/asy]$

We obviously see that we must use power of a point since they've given us lengths in a circle and there are intersection points. Let $T_1, T_2, T_3$ be our tangents from the circle to the parallelogram. By the secant power of a point, the power of $A = 3 cdot (3+9) = 36$ . Then $AT_2 = AT_3 = sqrt{36} = 6$ . Similarly, the power of $C = 16 cdot (16+9) = 400$ and $CT_1 = sqrt{400} = 20$ . We let $BT_3 = BT_1 = x$ and label the diagram accordingly.

Notice that because $BC = AD, 20+x = 6+DT_2 implies DT_2 = 14+x$ . Let $O$ be the center of the circle. Since $OT_1$ and $OT_2$ intersect $BC$ and $AD$ , respectively, at right angles, we have $T_2T_2CD$ is a right-angled trapezoid and more importantly, the diameter of the circle is the height of the triangle. Therefore, we can drop an altitude from $D$ to $BC$ and $C$ to $AD$ , and both are equal to $2r$ . Since $T_1E = T_2D$ , $20 - CE = 14+x implies CE = 6-x$ . Since $CE = DF, DF = 6-x$ and $AF = 6+14+x+6-x = 26$ . We can now use Pythagorean theorem on $triangle ACF$ ; we have $26^2 + (2r)^2 = (3+9+16)^2 implies 4r^2 = 784-676 implies 4r^2 = 108 implies 2r = 6sqrt{3}$ and $r^2 = 27$ .

We know that $CD = 6+x$ because $ABCD$ is a parallelogram. Using Pythagorean theorem on $triangle CDF$ , $(6+x)^2 = (6-x)^2 + 108 implies (6+x)^2-(6-x)^2 = 108 implies 12 cdot 2x = 108 implies 2x = 9 implies x = frac{9}{2}$ . Therefore, base $BC = 20 + frac{9}{2} = frac{49}{2}$ . Thus the area of the parallelogram is the base times the height, which is $frac{49}{2} cdot 6sqrt{3} = 147sqrt{3}$ and the answer is $boxed{150}$

~KingRavi

Solution 2

Let the circle tangent to $BC,AD,AB$ at $P,Q,M$ separately, denote that $angle{ABC}=angle{D}=alpha$

Using POP, it is very clear that $PC=20,AQ=AM=6$ , let $BM=BP=x,QD=14+x$ , using LOC in $triangle{ABP}$ , $x^2+(x+6)^2-2x(x+6)cosalpha=36+PQ^2$ , similarly, use LOC in $triangle{DQC}$ , getting that $(14+x)^2+(6+x)^2-2(6+x)(14+x)cosalpha=400+PQ^2$ . We use the second equation to minus the first equation, getting that $28x+196-(2x+12)*14*cosalpha=364$ , we can get $cosalpha=frac{2x-12}{2x+12}$ .

Now applying LOC in $triangle{ADC}$ , getting $(6+x)^2+400-2(6+x)*20*frac{2x-12}{2x+12}=(3+9+16)^2$ , solving this equation to get $x=frac{9}{2}$ , then $cosalpha=-frac{1}{7}$ , $sinalpha=frac{4sqrt{3}}{7}$ , the area is $frac{21}{2}*frac{49}{2}*frac{4sqrt{3}}{7}=147sqrt{3}$ leads to $boxed{150}$

~bluesoul

Solution 3

Denote by $O$ the center of the circle. Denote by $r$ the radius of the circle. Denote by $E$ , $F$ , $G$ the points that the circle meets $AB$ , $CD$ , $AD$ at, respectively.

Because the circle is tangent to $AD$ , $CB$ , $AB$ , $OE = OF = OG = r$ , $OE perp AD$ , $OF perp CB$ , $OG perp AB$ .

Because $AD parallel CB$ , $E$ , $O$ , $F$ are collinear.

Following from the power of a point, $AG^2 = AE^2 = AP cdot AQ$ . Hence, $AG = AE = 6$ .

Following from the power of a point, $CF^2 = CQ cdot CP$ . Hence, $CF = 20$ .

Denote $BG = x$ . Because $DG$ and $DF$ are tangents to the circle, $BF = x$ .

Because $AEFB$ is a right trapezoid, $AB^2 = EF^2 + left( AE - BF right)^2$ . Hence, $left( 6 + x right)^2 = 4 r^2 + left( 6 - x right)^2$ . This can be simplified as [ 6 x = r^2 . hspace{1cm} (1) ]

In $triangle ACB$ , by applying the law of cosines, we have begin{align*} AC^2 & = AB^2 + CB^2 - 2 AB cdot CB cos B \ & = AB^2 + CB^2 + 2 AB cdot CB cos A \ & = AB^2 + CB^2 + 2 AB cdot CB cdot frac{AE - BF}{AB} \ & = AB^2 + CB^2 + 2 CB left( AE - BF right) \ & = left( 6 + x right)^2 + left( 20 + x right)^2 + 2 left( 20 + x right) left( 6 - x right) \ & = 24 x + 676 . end{align*}

Because $AC = AP + PQ + QC = 28$ , we get $x = frac{9}{2}$ . Plugging this into Equation (1), we get $r = 3 sqrt{3}$ .

Therefore, begin{align*} {rm Area} ABCD & = CB cdot EF \ & = left( 20 + x right) cdot 2r \ & = 147 sqrt{3} . end{align*}

Therefore, the answer is $147 + 3 = boxed{textbf{(150) }}$ .

~Steven Chen (www.professorchenedu.com)

Solution 4

Let $omega$ be the circle, let $r$ be the radius of $omega$ , and let the points at which $omega$ is tangent to $AB$ , $BC$ , and $AD$ be $X$ , $Y$ , and $Z$ , respectively. Note that PoP on $A$ and $C$ with respect to $omega$ yields $AX=6$ and $CY=20$ . We can compute the area of $ABC$ in two ways:

1. By the half-base-height formula, $[ABC]=r(20+BX)$ .

2. We can drop altitudes from the center $O$ of $omega$ to $AB$ , $BC$ , and $AC$ , which have lengths $r$ , $r$ , and $sqrt{r^2-81/4}$ . Thus, $[ABC]=[OAB]+[OBC]+[OAC]=r(BX+13)+14sqrt{r^2-81/4}$ .

Equating the two expressions for $[ABC]$ and solving for $r$ yields $r=3sqrt{3}$ .

Let $BX=BY=a$ . By the Parallelogram Law, $(a+6)^2+(a+20)^2=38^2$ . Solving for $a$ yields $a=9/2$ . Thus, $[ABCD]=2[ABC]=2r(20+a)=147sqrt{3}$ , for a final answer of $boxed{150}$ .

~ Leo.Euler

Problem12

For any finite set $X$ , let $| X |$ denote the number of elements in $X$ . Define $[ S_n = sum | A cap B | , ]$ where the sum is taken over all ordered pairs $(A, B)$ such that $A$ and $B$ are subsets of $left{ 1 , 2 , 3, cdots , n right}$ with $|A| = |B|$ . For example, $S_2 = 4$ because the sum is taken over the pairs of subsets $[ (A, B) in left{ (emptyset, emptyset) , ( {1} , {1} ), ( {1} , {2} ) , ( {2} , {1} ) , ( {2} , {2} ) , ( {1 , 2} , {1 , 2} ) right} , ]$ giving $S_2 = 0 + 1 + 0 + 0 + 1 + 2 = 4$ . Let $frac{S_{2022}}{S_{2021}} = frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find the remainder when $p + q$ is divided by 1000.

Solution 1 (Easy to Understand)

Let's try out for small values of $n$ to get a feel for the problem. When $n=1, S_n$ is obviously $1$ . The problem states that for $n=2, S_n$ is $4$ . Let's try it out for $n=3$ .

Let's perform casework on the number of elements in $A, B$ .

$textbf{Case 1:} |A| = |B| = 1$

In this case, the only possible equivalencies will be if they are the exact same element, which happens $3$ times.

$textbf{Case 2:} |A| = |B| = 2$

In this case, if they share both elements, which happens $3$ times, we will get $2$ for each time, and if they share only one element, which also happens $6$ times, we will get $1$ for each time, for a total of $12$ for this case.

$textbf{Case 3:} |A| = |B| = 3$

In this case, the only possible scenario is that they both are the set ${1,2,3}$ , and we have $3$ for this case.

In total, $S_3 = 18$ .

Now notice, the number of intersections by each element $1 ldots 3$ , or in general, $1 ldots n$ is equal for each element because of symmetry - each element when $n=3$ adds $6$ to the answer. Notice that $6 = binom{4}{2}$ - let's prove that $S_n = n cdot binom{2n-2}{n-1}$ (note that you can assume this and answer the problem if you're running short on time in the real test).

Let's analyze the element $k$ - to find a general solution, we must count the number of these subsets that $k$ appears in. For $k$ to be in both $A$ and $B$ , we need $A = {k} cup A'| A' subset {1,2,ldots,n} land A' not subset {k}$ and $B = {k} cup B'| B' subset {1,2,ldots,n} land B' not subset {k}$ (Basically, both sets contain $k$ and another subset of $1$ through $n$ not including $k$ ).

For any $0<=l<=n-1$ that is the size of both $A'$ and $B'$ , the number of ways to choose the subsets $A'$ and $B'$ is $binom{n-1}{l}$ for both subsets, so the total number of ways to choose the subsets are $binom{n-1}{l}^2$ . Now we sum this over all possible $l$ 's to find the total number of ways to form sets $A$ and $B$ that contain $k$ . This is equal to $sum{l=0}{n-1} binom{n-1}{l}^2$ . This is a simplification of Vandermonde's identity, which states that $sum{k=0}{r} binom{m}{k} cdot binom{n}{r-k} = binom{m+n}{r}$ . Here, $m$ , $n$ and $r$ are all $n-1$ , so this sum is equal to $binom{2n-2}{n-1}$ . Finally, since we are iterating over all $k$ 's for $n$ values of $k$ , we have $S_n = n cdot binom{2n-2}{n-1}$ , proving our claim.

We now plug in $S_n$ to the expression we want to find. This turns out to be $frac{2022 cdot binom{4042}{2021}}{2021 cdot binom{4040}{2020}}$ . Expanding produces $frac{2022 cdot 4042!cdot 2020! cdot 2020!}{2021 cdot 4040! cdot 2021! cdot 2021!}$ .

After cancellation, we have $[frac{2022 cdot 4042 cdot 4041}{2021 cdot 2021 cdot 2021} implies frac{4044cdot 4041}{2021 cdot 2021}]$

$4044$ and $4041$ don't have any common factors with $2021$ , so we're done with the simplification. We want to find $4044 cdot 4041 + 2021^2 pmod{1000} = 44 cdot 41 + 21^2 pmod{1000} = 1804+481 pmod{1000} = 2245 pmod{1000} = boxed{245}$

~KingRavi

Linearity of Expectation Solution

We take cases based on the number of values in each of the subsets in the pair. Suppose we have $k$ elements in each of the subsets in a pair (for a total of n elements in the set). The expected number of elements in any random pair will be $n cdot frac{k}{n} cdot frac{k}{n}$ by linearity of expectation because for each of the $n$ elements, there is a $frac{k}{n}$ probability that the element will be chosen. To find the sum over all such values, we multiply this quantity by $binom{n}{k}^2$ . Summing, we get $[sum_{k=1}^{n} frac{k^2}{n} binom{n}{k}^2]$ Notice that we can rewrite this as $[sum_{k=1}^{n} frac{1}{n} left(frac{k cdot n!}{(k)!(n - k)!}right)^2 = sum_{k=1}^{n} frac{1}{n} n^2 left(frac{(n-1)!}{(k - 1)!(n - k)!}right)^2 = n sum_{k=1}^{n} binom{n - 1}{k - 1}^2 = n sum_{k=1}^{n} binom{n - 1}{k - 1}binom{n - 1}{n - k}]$ We can simplify this using Vandermonde's identity to get $n binom{2n - 2}{n - 1}$ . Evaluating this for $2022$ and $2021$ gives $[frac{2022binom{4042}{2021}}{2021binom{4040}{2020}} = frac{2022 cdot 4042 cdot 4041}{2021^3} = frac{2022 cdot 2 cdot 4041}{2021^2}]$ Evaluating the numerators and denominators mod $1000$ gives $804 + 441 = 1boxed{245}$

- pi_is_3.14

Solution 2 (Rigorous)

For each element $i$ , denote $x_i = left( x_{i, A}, x_{i, B} right) in left{ 0 , 1 right}^2$ , where $x_{i, A} = Bbb I left{ i in A right}$ (resp. $x_{i, B} = Bbb I left{ i in B right}$ ).

Denote $Omega = left{ (x_1, cdots , x_n): sum_{i = 1}^n x_{i, A} = sum_{i = 1}^n x_{i, B} right}$ .

Denote $Omega_{-j} = left{ (x_1, cdots , x_{j-1} , x_{j+1} , cdots , x_n): sum_{i neq j} x_{i, A} = sum_{i neq j} x_{i, B} right}$ .

Hence, $begin{align*} S_n & = sum_{(x_1, cdots , x_n) in Omega} sum_{i = 1}^n Bbb I left{ x_{i, A} = x_{i, B} = 1 right} \ & = sum_{i = 1}^n sum_{(x_1, cdots , x_n) in Omega} Bbb I left{ x_{i, A} = x_{i, B} = 1 right} \ & = sum_{i = 1}^n sum_{(x_1, cdots , x_{i-1} , x_{i+1} , cdots , x_n) in Omega_{-i}} 1 \ & = sum_{i = 1}^n sum_{j=0}^{n-1} left( binom{n-1}{j} right)^2 \ & = n sum_{j=0}^{n-1} left( binom{n-1}{j} right)^2 \ & = n sum_{j=0}^{n-1} binom{n-1}{j} binom{n-1}{n-1-j} \ & = n binom{2n-2}{n-1} . end{align*}$

Therefore, $begin{align*} frac{S_{2022}}{S_{2021}} & = frac{2022 binom{4042}{2021}}{2021 binom{4040}{2020}} \ & = frac{4044 cdot 4041}{2021^2} . end{align*}$

This is in the lowest term. Therefore, modulo 1000, $begin{align*} p + q & equiv 4044 cdot 4041 + 2021^2 \ & equiv 44 cdot 41 + 21^2 \ & equiv boxed{textbf{(245) }} . end{align*}$

~Steven Chen (www.professorchenedu.com

Problem13

Let $S$ be the set of all rational numbers that can be expressed as a repeating decimal in the form $0.overline{abcd},$ where at least one of the digits $a,$ $b,$ $c,$ or $d$ is nonzero. Let $N$ be the number of distinct numerators obtained when numbers in $S$ are written as fractions in lowest terms. For example, both $4$ and $410$ are counted among the distinct numerators for numbers in $S$ because $0.overline{3636} = frac{4}{11}$ and $0.overline{1230} = frac{410}{3333}.$ Find the remainder when $N$ is divided by $1000.$

Solution 1

$0.abcd=frac{overline{abcd}}{9999}$ , $9999=9times 11times 101$ .

Then we need to find the number of positive integers less than 10000 can meet the requirement.Suppose the number is x.

Case 1: (9999, x)=1. Clearly x satisfies. $[varphi left( 9999 right) =9999times left( 1-frac{1}{3} right) times left( 1-frac{1}{11} right) times left( 1-frac{1}{101} right)=6000]$

Case 2: 3|x but x is not a multiple of 11 or 101. Then the least value of abcd is 9x, so that $xle 1111$ , 334 values from 3 to 1110.

Case 3: 11|x but x is not a multiple of 3 or 101. Then the least value of abcd is 11x, so that $xle 909$ , 55 values from 11 to 902.

Case 4: 101|x. None.

Case 5: 3, 11|x. Then the least value of abcd is 11x, 3 values from 33 to 99.

To sum up, the answer is $[6000+334+55+3=boxed{6392}]$

Problem

Let $x,$ $y,$ and $z$ be positive real numbers satisfying the system of equations: $begin{align*} sqrt{2x-xy} + sqrt{2y-xy} &= 1 \ sqrt{2y-yz} + sqrt{2z-yz} &= sqrt2 \ sqrt{2z-zx} + sqrt{2x-zx} &= sqrt3. end{align*}$ Then $left[ (1-x)(1-y)(1-z) right]^2$ can be written as $frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution 1 (geometric interpretation)

First, we note that we can let a triangle exist with side lengths $sqrt{2x}$ , $sqrt{2z}$ , and opposite altitude $sqrt{xz}$ . This shows that the third side, which is the nasty square-rooted sum, is going to have the length equal to the sum on the right - let this be $l$ for symmetry purposes. So, we note that if the angle opposite the side with length $sqrt{2x}$ has a value of $sin(theta)$ , then the altitude has length $sqrt{2z} cdot sin(theta) = sqrt{xz}$ and thus $sin(theta) = sqrt{frac{x}{2}}$ so $x=2sin^2(theta)$ and the triangle side with length $sqrt{2x}$ is equal to $2sin(theta)$ .

We can symmetrically apply this to the two other triangles, and since by law of sines, we have $frac{2sin(theta)}{sin(theta)} = 2R to R=1$ is the circumradius of that triangle. Hence. we calculate that with $l=1, sqrt{2}$ , and $sqrt{3}$ , the angles from the third side with respect to the circumcenter are $120^{circ}, 90^{circ}$ , and $60^{circ}$ . This means that by half angle arcs, we see that we have in some order, $x=2sin^2(alpha)$ , $x=2sin^2(beta)$ , and $z=2sin^2(gamma)$ (not necessarily this order, but here it does not matter due to symmetry), satisfying that $alpha+beta=180^{circ}-frac{120^{circ}}{2}$ , $beta+gamma=180^{circ}-frac{90^{circ}}{2}$ , and $gamma+alpha=180^{circ}-frac{60^{circ}}{2}$ . Solving, we get $alpha=frac{135^{circ}}{2}$ , $beta=frac{105^{circ}}{2}$ , and $gamma=frac{165^{circ}}{2}$ .

We notice that $[[(1-x)(1-y)(1-z)]^2=[sin(2alpha)sin(2beta)sin(2gamma)]^2=[sin(135^{circ})sin(105^{circ})sin(165^{circ})]^2]$ $[=left(frac{sqrt{2}}{2} cdot frac{sqrt{6}-sqrt{2}}{4} cdot frac{sqrt{6}+sqrt{2}}{4}right)^2 = left(frac{sqrt{2}}{8}right)^2=frac{1}{32} to boxed{033}. blacksquare]$

- kevinmathz

Solution 2 (pure algebraic trig, easy to follow)

(This eventually whittles down to the same concept as Solution 1)

Note that in each equation in this system, it is possible to factor $sqrt{x}$ , $sqrt{y}$ , or $sqrt{z}$ from each term (on the left sides), since each of $x$ , $y$ , and $z$ are positive real numbers. After factoring out accordingly from each terms one of $sqrt{x}$ , $sqrt{y}$ , or $sqrt{z}$ , the system should look like this: $begin{align*} sqrt{x}cdotsqrt{2-y} + sqrt{y}cdotsqrt{2-x} &= 1 \ sqrt{y}cdotsqrt{2-z} + sqrt{z}cdotsqrt{2-y} &= sqrt2 \ sqrt{z}cdotsqrt{2-x} + sqrt{x}cdotsqrt{2-z} &= sqrt3. end{align*}$ This should give off tons of trigonometry vibes. To make the connection clear, $x = 2cos^2 alpha$ , $y = 2cos^2 beta$ , and $z = 2cos^2 theta$ is a helpful substitution: $begin{align*} sqrt{2cos^2 alpha}cdotsqrt{2-2cos^2 beta} + sqrt{2cos^2 beta}cdotsqrt{2-2cos^2 alpha} &= 1 \ sqrt{2cos^2 beta}cdotsqrt{2-2cos^2 theta} + sqrt{2cos^2 theta}cdotsqrt{2-2cos^2 beta} &= sqrt2 \ sqrt{2cos^2 theta}cdotsqrt{2-2cos^2 alpha} + sqrt{2cos^2 alpha}cdotsqrt{2-2cos^2 theta} &= sqrt3. end{align*}$ From each equation $sqrt{2}^2$ can be factored out, and when every equation is divided by 2, we get: $begin{align*} sqrt{cos^2 alpha}cdotsqrt{1-cos^2 beta} + sqrt{cos^2 beta}cdotsqrt{1-cos^2 alpha} &= frac{1}{2} \ sqrt{cos^2 beta}cdotsqrt{1-cos^2 theta} + sqrt{cos^2 theta}cdotsqrt{1-cos^2 beta} &= frac{sqrt2}{2} \ sqrt{cos^2 theta}cdotsqrt{1-cos^2 alpha} + sqrt{cos^2 alpha}cdotsqrt{1-cos^2 theta} &= frac{sqrt3}{2}. end{align*}$ which simplifies to (using the Pythagorean identity $sin^2 phi + cos^2 phi = 1 ; forall ; phi in mathbb{C}$ ): $begin{align*} cos alphacdotsin beta + cos betacdotsin alpha &= frac{1}{2} \ cos betacdotsin theta + cos thetacdotsin beta &= frac{sqrt2}{2} \ cos thetacdotsin alpha + cos alphacdotsin theta &= frac{sqrt3}{2}. end{align*}$ which further simplifies to (using sine addition formula $sin(a + b) = sin a cos b + cos a sin b$ ): $begin{align*} sin(alpha + beta) &= frac{1}{2} \ sin(beta + theta) &= frac{sqrt2}{2} \ sin(alpha + theta) &= frac{sqrt3}{2}. end{align*}$ Without loss of generality, taking the inverse sine of each equation yields a simple system: $begin{align*} alpha + beta &= frac{pi}{6} \ beta + theta &= frac{pi}{4} \ alpha + theta &= frac{pi}{3}. end{align*}$ giving solutions $alpha = frac{pi}{8}$ , $beta = frac{pi}{24}$ , $theta = frac{5pi}{24}$ . Since these unknowns are directly related to our original unknowns, there are consequent solutions for those: $x = 2cos^2left(frac{pi}{8}right)$ , $y = 2cos^2left(frac{pi}{24}right)$ , and $z = 2cos^2left(frac{5pi}{24}right)$ . When plugging into the expression $left[ (1-x)(1-y)(1-z) right]^2$ , noting that $-cos 2phi = 1 - 2cos^2 phi; forall ; phi in mathbb{C}$ helps to simplify this expression into:

Now, all the cosines in here are fairly standard: $cos frac{pi}{4} = frac{sqrt{2}}{2}$ , $;$ $cos frac{pi}{12} = frac{sqrt{6} + sqrt{2}}{4}$ , $;$ and $cos frac{5pi}{12} = frac{sqrt{6} - sqrt{2}}{4}$ . With some final calculations: $[(-1)^2left(frac{sqrt{2}}{2}right)^2left(frac{sqrt{6} + sqrt{2}}{4}right)^2left(frac{sqrt{6} - sqrt{2}}{4}right)^2 = left(frac{1}{2}right)left(frac{2 + sqrt{3}}{4}right)left(frac{2 - sqrt{3}}{4}right) = frac{left(2 - sqrt{3}right)left(2 + sqrt{3}right)}{2cdot4cdot4} = frac{1}{32}.]$ This is our answer in simplest form $frac{m}{n}$ , so $m + n = 1 + 32 = boxed{033}.$