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2018 AMC 10A Answer Key答案解析

Category: AMC美国数学竞赛, 国际竞赛, 福利干货, 竞赛资料 Date: 2018年2月9日上午10:23

2018 AMC 10A Answer Key答案解析

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Problem 1

答案：B
Solution： $2+1=3$ , so the reciprocal is $\frac{1}{3}$ . Summing $1$ and repeating leads to $\frac{3}{4}$ . Rinse and repeat to get $\frac{4}{7}$ . Adding $1$ to that is $\frac{11}{7}$ (B)

Problem 2

答案：A
Solution：Let's assume that Jacqueline has $1$ gallon of soda. Then Alice has $1.25$ gallons and Liliane has $1.5$ gallons. Doing division, we find out that $\frac{1.5}{1.25}=1.2$ , which means that Liliane has $20\%$ more soda. Therefore, the answer is $\boxed{\textbf{(A)}}$

Problem 3

答案：E
Solution 1：There are $10!$ seconds that the blood has before expiring and there are $60 \cdot 60 \cdot 24$ seconds in a day. Dividing $\frac{10!}{60 \cdot 60 \cdot 24}$ gives $42$ days. $42$ days after January 1 is $\fbox{\textbf{(E) }\text{February 12}}$ .
Solution 2： $10! = 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1$ . Convert $10!$ seconds to minutes: $9\cdot 8\cdot 7\cdot 5\cdot 4\cdot 3\cdot 2$ minutes. Convert minutes to hours: $9\cdot 8\cdot 7\cdot 2$ hours. Convert hours to days: $3\cdot 7\cdot 2 = 42$ days. $42$ days after January 1 is $\fbox{\textbf{(E) }\text{February 12}}$

Problem 4

答案：E
Solution 1：

We must place the classes into the periods such that no two balls are in the same period or in consecutive period. Ignoring distinguishability, we can thus list out the ways that three periods can be chosen for the classes, when periods cannot be consecutive:

Periods $1, 3, 5$

Periods $1, 3, 6$

Periods $1, 4, 6$

Periods $2, 4, 6$

There are $4$ ways to place $3$ nondistinguishable classes into $6$ periods such that no two classes are in consecutive periods. For each of these ways, there are $3! = 6$ orderings of the classes among themselves. Therefore, there are $4 \cdot 6 = \boxed{\mathrm{(E) \ } 24}$ ways to choose the classes.

Solution 2：

First draw 6 $X$ 's representing the 6 periods.

$XXXXXX$

Let the $O$ 's represent the classes that occupy each period.

$XOXOXO$

There are 6 ways to place the first class. There are 4 ways to place the second class. There is 1 way to place the third class. We multiply $6 \cdot 4 \cdot 1= \boxed{(E) 24}$ Solution by Baolan

Problem 5：

答案：D
Solution：From Alice and Bob, we know that $5 < d < 6.$ From Charlie, we know that $4 < d.$ We take the union of these two intervals to yield $\boxed{\textbf{(D)} (5,6)}.$

Problem 6：

答案：B
Solution：If $65\%$ of the votes were likes, then $35\%$ of the votes were dislikes. $65\%-35\%=30\%$ , so $90$ votes is $30\%$ of the total number of votes. Doing quick arithmetic shows that the answer is $300\Rightarrow\text{\boxed{B}}$

Problem 7：

答案：E
Solution：The prime factorization of $4000$ is $2^{5}5^{3}$ . Therefore, the maximum number for $n$ is $3$ , and the minimum number for $n$ is $-5$ . The range from $-5$ to $3$ , which is $3-(-5) + 1 = 8 + 1 = \fbox{\textbf{(E) }9}$ .

Problem 8：

答案：C
Solution 1：Let $x$ be the number of 5-cent coins that Joe has. Therefore, he must have $(x+3)$ 10-cent coins and $(23-(x+3)-x)$ 25-cent coins. Since the total value of his collection is 320 cents, we can write $5x+10(x+3)+25(23-(x+3)-x)= 320 \Rightarrow 5x+10x+30+500-50x= 320 \Rightarrow 35x= 210 \Rightarrow x= 6$ Joe has 6 5-cent coins, 9 10-cent coins, and 8 25-cent coins. Thus, our answer is $8-6=\boxed{2}$
Solution 2：Let n be the number of 5 cent coins Joe has, d be the number of 10 cent coins, and q the number of 25 cent coins. We are solving for q - n. We know that the value of the coins add up to 320 cents. Thus, we have 5n + 10d + 25q = 320. Let this be (1). We know that there are 23 coins. Thus, we have n + d + q = 23. Let this be (2). We know that there are 3 more dimes than nickels, which also means that there are 3 less nickels than dimes. Thus, we have d - 3 = n. Plugging d-3 into the other two equations for n, (1) becomes 2d + q - 3 = 23 and (2) becomes 15d + 25q - 15 = 320. (1) then becomes 2d + q = 26, and (2) then becomes 15d + 25q = 335. Multiplying (1) by 25, we have 50d + 25q = 650 (or 25^2 + 25). Subtracting (2) from (1) gives us 35d = 315, which means d = 9. Plugging d into d - 3 = n, n = 6. Plugging d and q into the (2) we had at the beginning of this problem, q = 8. Thus, the answer is 8 - 6 = 2.

Problem 9：

答案：E
Solution 1：Let $x$ be the area of $ADE$ . Note that $x$ is comprised of the $7$ small isosceles triangles and a triangle similar to $ADE$ with side length ratio $3:4$ (so an area ratio of $9:16$ ). Thus, we have $x=7+\dfrac{9}{16}x$ This gives $x=16$ , so the area of $DBCE=40-x=\boxed{24}$ .
Solution 2：The area of $ADE$ is 16 times the area of the small triangle, as they are similar and their side ratio is $4:1$ . Therefore the area of the trapezoid is $40-16= \boxed{24}$ .

Problem 10：

答案：A
Solution 1： $(\sqrt {49-x^2} + \sqrt {25-x^2}) * (\sqrt {49-x^2} - \sqrt {25-x^2}) = 49-x^2 - 25 +x^2 = 24$ ，Given that $(\sqrt {49-x^2} - \sqrt {25-x^2})$ = 3, $(\sqrt {49-x^2} + \sqrt {25-x^2}) = \frac {24} {3} = \boxed{(A) 8}$ Solution by PancakeMonster2004.
Solution 2：Let $u=\sqrt{49-x^2}$ , and let $v=\sqrt{25-x^2}$ . Then $v=\sqrt{u^2-24}$ . Substituting, we get $u-\sqrt{u^2-24}=3$ . Rearranging, we get $u-3=\sqrt{u^2-24}$ . Squaring both sides and solving, we get $u=\frac{11}{2}$ and $v=\frac{11}{2}-3=\frac{5}{2}$ . Adding, we get that the answer is $\boxed{(A) 8}$

Problem 11：

答案：E
Solution 1：The minimum number that can be shown on the face of a die is 1, so the least possible sum of the top faces of the 7 dies is 7. In order for the sum to be exactly 10, 1-3 dices' number on the top face must be increased by a total of 3. There are 3 ways to do so: 3, 2+1, and 1+1+1.There are $\dbinom {7}{1}$ for Case 1, $7*6 = 42$ for Case 2, and $\dbinom {7}{3}$ for Case 3. Therefore, the answer is $7+42+35 = \boxed {(E) 84}$ . Solution by PancakeMonster2004
Solution 2: Rolling a sum of 10 with 7 dice can be represented with stars and bars, with 10 stars and 6 bars. Each star represents one of the dots on the die's faces and the bars represent separation between different dice. However, we must note that each die must have at least one dot on a face, so there must already be 7 stars predetermined. We are left with 3 stars and 6 bars, which we can rearrange in $\dbinom{9}{3}=\boxed{84}$ ways. By RegularHexagon

Problem 12：

答案：C
Solution 1：

The graph looks something like this:

$[asy] draw((-3,0)--(3,0), Arrows); draw((0,-3)--(0,3), Arrows); draw((2,3)--(0,1)--(-2,3), blue); draw((-3,2)--(-1,0)--(-3,-2), blue); draw((-2,-3)--(0,-1)--(2,-3), blue); draw((3,-2)--(1,0)--(3,2), blue); draw((-3,2)--(3,0), red); dot((-3,2)); dot((3,0)); dot((0,1)); [/asy]$

Now it's clear that there are $\boxed{3}$ intersection points. (pinetree1)

Solution 2:

$x+3y=3$ can be rewritten to $x=3-3y$ . Substituting $3-3y$ for $x$ in the second equation will give $||3-3y|-y|=1$ . Splitting this question into casework for the ranges of y will give us the total number of solutions.

Case 1: $y>1$

$3-3y$ will be negative so $|3-3y| = 3y-3$

$|3y-3-y| = |2y-3| = 1$

   Subcase 1:

$2y-3$ is positive so $2y-3 = 1$ and $y = 2$ and $x = 3-3(2) = -3$

   Subcase 2:

$2y-3$ is negative so $|2y-3| = 3-2y = 1$ . $2y = 2$ and so there are no solutions ( $y$ can't equal to $1$ )

Case 2: $y = 1 x = 0$

Case 3: $y<1$

$3-3y$ will be positive so $|3-3y-y| = |3-4y| = 1$

   Subcase 1:

$3-4y$ will be negative so $4y-3 = 1$ --> $4y = 4$ . There are no solutions (again, $y$ can't equal to $1$ )

   Subcase 2: y<4/3

$3-4y$ will be positive so $3-4y = 1$ --> $4y = 2$ . $y = \frac{1}{2}$ and $x = \frac{3}{2}$ Solutions: $(-3,2) (0,1) (\frac{3}{2},\frac{1}{2})$ $\boxed{\textbf{(C)} \ 3}$

NOTE: Please fix this up using latex I have no idea how

Solution by Danny Li JHS

Problem 13

答案：D
Solution 1：First, we need to realize that the crease line is just the perpendicular bisector of side $AB$ , the hypotenuse of right triangle $\triangle ABC$ . Call the midpoint of $AB$ point $D$ . Draw this line and call the intersection point with $AC$ as $E$ . Now, $\triangle ABC$ is similar to $\triangle ADE$ by $AA$ similarity. Setting up the ratios, we find that $\frac{BC}{AC}=\frac{DE}{AD} \Rightarrow \frac{3}{4}=\frac{DE}{\frac{5}{2}} \Rightarrow DE=\frac{15}{8}.$ Thus, our answer is $\boxed{D}$ . Solution by Nivek
Solution 2 (if you are already out of time)：Simply make a 3x4x5 inch triangle and then cut it out (using fine rips). Then, make the fold and measure. It will be $\boxed{D} \frac{15}{8}$ inches in length.

Problem 14

答案：A
Solution 1：Let's set this value equal to $x$ . We can write $\frac{3^{100}+2^{100}}{3^{96}+2^{96}}=x.$ Multiplying by $3^{96}+2^{96}$ on both sides, we get $3^{100}+2^{100}=x(3^{96}+2^{96}).$ Now let's take a look at the answer choices. We notice that $81$ , choice $B$ , can be written as $3^4$ . Plugging this into out equation above, we get $3^{100}+2^{100} \stackrel{?}{=} 3^4(3^{96}+2^{96}) \Rightarrow 3^{100}+2^{100} \stackrel{?}{=} 3^{100}+3^4*2^{96}.$ The right side is larger than the left side because $2^{100} \leq 2^{96}*3^4.$ This means that our original value, $x$ , must be less than $81$ . The only answer that is less than $81$ is $80$ so our answer is $\boxed{A}$ . Solution by Nivek
Solution 2：Let $x=3^{96}$ and $y=2^{96}$ . Then our fraction can be written as $\frac{81x+16y}{x+y}=\frac{16x+16y}{x+y}+\frac{65x}{x+y}=16+\frac{65x}{x+y}$ . Notice that $\frac{65x}{x+y}<\frac{65x}{x}=65$ . So , $16+\frac{65x}{x+y}<16+65=81$ . And our only answer choice less than 81 is $\boxed{(A)}$ Solution by RegularHexagon
Solution 3：Let $x=\frac{3^{100}+2^{100}}{3^{96}+2^{96}}$ . Multiply both sides by $(3^{96}+2^{96})$ , and expand. Rearranging the terms, we get $3^{96}(3^4-x)+2^{96}(2^4-x)=0$ . The left side is strictly decreasing, and it is negative when $x=81$ . This means that the answer must be less than $81$ ; therefore the answer is $\boxed{(A)}$ .
Solution 4：A faster solution. Recognize that for exponents of this size $3^{n}$ will be enormously greater than $2^{n}$ , so the terms involving $2$ will actually have very little effect on the quotient. Now we know the answer will be very close to $81$ . Notice that the terms being added on to the top and bottom are in the ratio $\frac{1}{16}$ with each other, so they must pull the ratio down from 81 very slightly. (In the same way that a new test score lower than your current cumulative grade always must pull that grade downward.) Answer: $\boxed{(A)}$ .

Problem 15

答案：D
Solution 1 (PLEASE DRAW A PICTURE) ：Call center of the largest circle $X$ . The circle that is tangent at point $A$ will have point $Y$ as the center. Similarly, the circle that is tangent at point $B$ will have point $Z$ as the center. Connect $AB$ , $YZ$ , $XA$ , and $WY$ . Now observe that $\triangle XYZ$ is similar to $\triangle XAB$ . Writing out the ratios, we get $\frac{XY}{XA}=\frac{YZ}{AB} \Rightarrow \frac{13-5}{13}=\frac{5+5}{AB} \Rightarrow \frac{8}{13}=\frac{10}{AB} \Rightarrow AB=\frac{65}{4}.$ Therefore, our answer is $65+4=$ $\boxed{69}$ , which is choice $\boxed{D}$ .
Solution 2：

$[asy] draw(circle((0,0),13)); draw(circle((5,-6.25),5)); draw(circle((-5,-6.25),5)); label("$A$", (-8.125,-10.15), S); label("$B$", (8.125,-10.15), S); label("$C$", (0,-6.25), NE); draw((0,0)--(-8.125,-10.15)); draw((0,0)--(8.125,-10.15)); draw((-5,-6.25)--(5,-6.25)); draw((0,0)--(0,-13)); draw((-8.125,-10.15)--(8.125,-10.15)); label("$O$", (0,0), N); [/asy]$ Let the center of the large circle be $O$ . Let the common tangent of the two smaller circles be $C$ . Draw the two radii of the large circle, $\overline{OA}$ and $\overline{OB}$ and the two radii of the smaller circles to point $C$ . Draw ray $\overrightarrow{OC}$ . Draw $\overline{AB}$ . This sets us up with similar triangles, which we can solve. The length of $\overline{OC}$ is equal to $\sqrt{39}$ by Pythagorean Theorem, the length of the hypotenuse is 8, and the other leg is 5. Using similar triangles, $OB$ is 13, and therefore half of $AB$ is $\frac{65}{8}$ . Doubling gives $\frac{65}{4}$ which results in $65+4=\boxed{69}$ , which is choice $\boxed{D}$ .

Problem 16

答案：D
Solution：The hypotenuse has length $29$ . Let $P$ be the foot of the altitude from $B$ to $AC$ . Note that $BP$ is the shortest possible length of any segment. Writing the area of the triangle in two ways, we can solve for $BP=\dfrac{20\cdot 21}{29}$ , which is between $14$ and $15$ . Let the line segment be $BX$ , with $X$ on $AC$ . As you move $X$ along the hypotenuse from $A$ to $P$ , the length of $BX$ strictly decreases, hitting all the integer values from $20, 19, \dots 15$ . Similarly, moving $X$ from $P$ to $C$ hits all the integer values from $15, 16, \dots, 21$ . This is a total of $\boxed{13}$ line segments.

Problem 17

答案：C
Solution：

If we start with $1$ , we can include nothing else, so that won't work.

If we start with $2$ , we would have to include every odd number except $1$ to fill out the set, but then $3$ and $9$ would violate the rule, so that won't work.

Experimentation with $3$ shows it's likewise impossible. You can include $7$ , $11$ , and either $5$ or $10$ (which are always safe). But after adding either $4$ or $8$ we have nowhere else to go.

Finally, starting with $4$ , we find that the sequence $4,5,6,7,9,11$ works, giving us $\boxed{\textbf{(C)} \text{ 4}}$ . (Random_Guy)

Problem 18：

答案：D
Solution 1：This looks like balanced ternary, in which all the integers with absolute values less than $\frac{3^n}{2}$ are represented in $n$ digits. There are 8 digits. Plugging in 8 into the formula gives a maximum bound of $|x|=3280.5$ , which means there are 3280 positive integers, 0, and 3280 negative integers. Since we want all nonnegative integers, there are $3280+1=\boxed{3281}$ integers or $\boxed{D}$ .
Solution 2：Note that all numbers formed from this sum are either positive, negative or zero. The number of positive numbers formed by this sum is equal to the number of negative numbers formed by this sum, because of symmetry. There is only one way to achieve a sum of zero, if all $a_i=0$ . The total number of ways to pick $a_i$ from $i=1, 2, 3, ... 7$ is $3^8=6561$ . $\frac{6561-1}{2}=3280$ gives the number of possible negative integers. The question asks for the number of nonnegative integers, so subtracting from the total gives $6561-3280=\boxed{3281}$ .Solution by RegularHexagon
Solution 3 (Quick Solution)：Note that the number of total possibilities (ignoring the conditions set by the problem) is $3^8=6561$ . So, E is clearly unrealistic. Note that if $a_7$ is 1, then it's impossible for $a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,$ to be negative. Therefore, if $a_7$ is 1, there are $3^7=2187$ possibilities. As A, B, and C are all less than 2187, the answer must be $\boxed{(D) 3281}$

Problem 19

答案：E
Solution 1：

Since we only care about the unit digit, our set $\{11,13,15,17,19 \}$ can be turned into $\{1,3,5,7,9 \}$ . Call this set $A$ and call $\{1999, 2000, 2001, \cdots , 2018 \}$ set $B$ . Let's do casework on the element of $A$ that we choose. Since $1*1=1$ , any number from $B$ can be paired with $1$ to make $1^n$ have a units digit of $1$ . Therefore, the probability of this case happening is $\frac{1}{5}$ since there is a $\frac{1}{5}$ chance that the number $1$ is selected from $A$ . Let us consider the case where the number $3$ is selected from $A$ . Let's look at the unit digit when we repeatedly multiply the number $3$ by itself:

$3*3=9$

$9*3=7$

$7*3=1$

$1*3=3$

We see that the unit digit of $3^x$ , for some integer $x$ , will only be $1$ when $x$ is a multiple of $4$ . Now, let's count how many numbers in $B$ are divisible by $4$ . This can be done by simply listing: $2000,2004,2008,2012,2016.$ There are $5$ numbers in $B$ divisible by $4$ out of the $2018-1999+1=20$ total numbers. Therefore, the probability that $3$ is picked from $A$ and a number divisible by $4$ is picked from $B$ is $\frac{1}{5}*\frac{5}{20}=\frac{1}{20}$ . Similarly, we can look at the repeating units digit for $7$ :

$7*7=9$

$9*7=3$

$3*7=1$

$1*7=7$

We see that the unit digit of $7^y$ , for some integer $y$ , will only be $1$ when $y$ is a multiple of $4$ . This is exactly the same conditions as our last case with $3$ so the probability of this case is also $\frac{1}{20}$ . Since $5*5=25$ and $25$ ends in $5$ , the units digit of $5^w$ , for some integer, $w$ will always be $5$ . Thus, the probability in this case is $0$ . The last case we need to consider is when the number $9$ is chosen from $A$ . This happens with probability $\frac{1}{5}$ . We list out the repeating units digit for $9$ as we have done for $3$ and $7$ :

$9*9=1$

$1*9=9$

We see that the units digit of $9^z$ , for some integer $z$ , is $1$ only when $z$ is an even number. From the $20$ numbers in $B$ , we see that exactly half of them are even. The probability in this case is $\frac{1}{5}*\frac{1}{2}=\frac{1}{10}.$ Finally, we can add all of our probabilities together to get $\frac{1}{5}+\frac{1}{20}+\frac{1}{20}+\frac{1}{10}=\boxed{\frac{2}{5}}.$

~Nivek

Solution 2：Since only the units digit is relevant, we can turn the first set into $\{1,3,5,7,9\}$ . Note that $x^4 \equiv 1 \mod 10$ for all odd digits $x$ , except for 5. Looking at the second set, we see that it is a set of all integers between 1999 and 2018. There are 20 members of this set, which means that, $\mod 4$ , this set has 5 values which correspond to $\{0,1,2,3\}$ , making the probability equal for all of them. Next, check the values for which it is equal to $1 \mod 10$ . There are $4+1+0+1+2=8$ values for which it is equal to 1, remembering that $5^{4n} \equiv 1 \mod 10$ only if $n=0$ , which it is not. There are 20 values in total, and simplifying $\frac{8}{20}$ gives us $\boxed{\frac{2}{5}}$ or $\boxed{E}$ .

Problem 20

答案：B
Solution：

Draw a $7 \times 7$ square.

$\begin{tabular}{|c|c|c|c|c|c|c|} \hline a & b & c & d & c & b & a \\ \hline b & e & f & g & f & e & b \\ \hline c & f & h & i & h & f & c \\ \hline d & g & i & j & i & g & d \\ \hline c & f & h & i & h & f & c \\ \hline b & e & f & g & f & e & b \\ \hline a & b & c & d & c & b & a \\ \hline \end{tabular}$

There are only ten squares we get to actually choose, and two independent choices for each, for a total of $2^{10} = 1024$ codes. Two codes must be subtracted (due to the rule that there must be at least one square of each color) for an answer of $\fbox{\textbf{(B)} \text{ 1022}}$

Problem 21

答案：E
Solution 1：

Substituting $y=x^2-a$ into $x^2+y^2=a^2$ , we get $x^2+(x^2-a)^2=a^2 \implies x^2+x^4-2ax^2=0 \implies x^2(x^2-(2a-1))=0$ Since this is a quartic, there are 4 total roots (counting multiplicity). We see that $x=0$ always at least one intersection at $(0,-a)$ (and is in fact a double root). The other two intersection points have $x$ coordinates $\sqrt{2a-1}$ . We must have $2a-1> 0,$ otherwise we are in the case where the parabola lies entirely above the circle (tangent to it at the point $(0,a)$ ). This only results in a single intersection point in the real coordinate plane. Thus, we see $a>\frac{1}{2}$ . (projecteulerlover)

Solution 2：

Looking at a graph, it is obvious that the two curves intersect at (0, -a). We also see that if the parabola go's 'in' the circle, than by going out of it (as it will) it will intersect five times, an impossibility. Thus we only look for cases where the parabola becomes externally tangent to the circle. We have $x^2 - a = -\sqrt(a^2 - x^2)$ . Squaring both sides and solving yields $x^4 - (2a - 1)x^2 = 0$ . Since x = 0 is already accounted for, we only need to find 1 solution for $x^2 = 2a - 1$ , where the right hand side portion is obviously increasing. Since a = 1/2 begets x = 0 (an overcount), we have $a > 1/2 -> E$ is the right answer. Solution by JohnHankock

Solution 3：

This describes a unit parabola, with a circle centered at the axis of symmetry and tangent to the vertex. As the curvature of the unit parabola at the vertex is 2, the radius of the circle that matches it has a radius of $\frac{1}{2}$ . This circle is tangent to an infinitesimally close pair of points, one on each side. Therefore, it is tangent to only 1 point. When a larger circle is used, it is tangent to 3 points because the points on either side are now separated from the vertex. Therefore, $\boxed{a > \frac{1}{2}}$ or $\boxed{E}$ is correct.

Solution 4：

Notice, the equations are of that of a circle of radius a centered at the origin and a parabola translated down by a units. They always intersect at the point $(0, a)$ , and they have symmetry across the y-axis, thus, for them to intersect at exactly 3 points, it suffices to find the y solution.

First, rewrite the second equation to $y=x^2-a\implies x^2=y+a$ And substitute into the first equation: $y+a+y^2=a^2$ Since we're only interested in seeing the interval in which a can exist, we find the discriminant: $1-4a+4a^2$ . This value must not be less than 0 (It is the square root part of the quadratic formula). To find when it is 0, we find the roots: $4a^2-4a+1=0 \implies a=\frac{4\pm\sqrt{16-16}}{8}=\frac{1}{2}$ Since $\lim_{a\to \infty}(4a^2-4a+1)=\infty$ , our range is $\boxed{a>\frac{1}{2}}$

Problem 22

答案：D
Solution：We can say that $a$ and $b$ 'have' $2^3 * 3$ , that $b$ and $c$ have $2^2 * 3^2$ , and that $c$ and $d$ have $3^3 * 2$ . Combining $1$ and $2$ yields $b$ has (at a minimum) $2^3 * 3^2$ , and thus $a$ has $2^3 * 3$ (and no more powers of $3$ because otherwise $gcd(a,b)$ would be different). In addition, $c$ has $3^3 * 2^2$ , and thus $d$ has $3^3 * 2$ (similar to $a$ , we see that $d$ cannot have any other powers of $2$ ). We now assume the simplest scenario, where $a = 2^3 * 3$ and $d = 3^3 * 2$ . According to this base case, we have $gcd(a, d) = 2 * 3 = 6$ . We want an extra factor between the two such that this number is between $70$ and $100$ , and this new factor cannot be divisible by $2$ or $3$ . Checking through, we see that $6 * 13$ is the only one that works. Therefore the answer is $\boxed{D}$ Solution by JohnHankock

Problem 23

答案：D
Solution 1： Let the square have side length $x$ . Connect the upper-right vertex of square $S$ with the two vertices of the triangle's hypotenuse. This divides the triangle in several regions whose areas must add up to the area of the whole triangle, which is $6$ . Square $S$ has area $x^2$ , and the two thin triangle regions have area $\dfrac{x(3-x)}{2}$ and $\dfrac{x(4-x)}{2}$ . The final triangular region with the hypotenuse as its base and height $2$ has area $5$ . Thus, we have $x^2+\dfrac{x(3-x)}{2}+\dfrac{x(4-x)}{2}+5=6$ Solving gives $x=\dfrac{2}{7}$ . The area of $S$ is $\dfrac{4}{49}$ and the desired ratio is $\dfrac{6-\dfrac{4}{49}}{6}=\boxed{\dfrac{145}{147}}$ .

Solution 2: Let the square have side length $s$ . If we were to extend the sides of the square further into the triangle until they intersect on point on the hypotenuse, we'd have a similar right triangle formed between the hypotenuse and the two new lines, and 2 smaller similar triangles that share a side of length 2. Using the side-to-side ratios of these triangles, we can find that the length of the big similar triangle is $\frac{5}{3}(2)=\frac{10}{3}$ . Now, let's extend this big similar right triangle to the left until it hits the side of length 3. Now, the length is $\frac{10}{3}+s$ , and using the ratios of the side lengths, the height is $\frac{3}{4}(\frac{10}{3}+s)=\frac{5}{2}+\frac{3s}{4}$ . Looking at the diagram, if we add the height of this triangle to the side length of the square, we'd get 3, so $\frac{5}{2}+\frac{3s}{4}+s=\frac{5}{2}+\frac{7s}{4}=3 \\ \frac{7s}{4}=\frac{1}{2} \\ s=\frac{2}{7} \implies \textrm{ area of square is } (\frac{2}{7})^2=\frac{4}{49}$ Now comes the easy part: finding the ratio of the areas: $\frac{3\cdot 4 \cdot \frac{1}{2} -\frac{4}{49}}{3\cdot 4 \cdot \frac{1}{2}}=\frac{6-\frac{4}{49}}{6}=\frac{294-4}{294}=\frac{290}{294}=\boxed{\frac{145}{147}.}$

Solution 3: We use coordinate geometry. Let the right angle be at $(0,0)$ and the hypotenuse be the line $3x+4y = 12$ for $0\le x\le 3$ . Denote the position of $S$ as $(s,s)$ , and by the point to line distance formula, we know that $\frac{|3s+4s-12|}{5} = 2$ $\Rightarrow |7s-12| = 10$ Obviously $s<\frac{22}{7}$ , so $s = \frac{2}{7}$ , and from here the rest of the solution follows to get $\boxed{\frac{145}{147}}$ .
Solution 4: Let the side length of the square be $x$ . First off, let us make a similar triangle with the segment of length $2$ and the top-right corner of $S$ . Therefore, the longest side of the smaller triangle must be $2 \cdot \frac54 = \frac52$ . We then do operations with that side in terms of $x$ . We subtract $x$ from the bottom, and $\frac{3x}{4}$ from the top. That gives us the equation of $3-\frac{7x}{4} = \frac{5}{2}$ . Solving, $12-7x = 10 \implies x = \frac{2}{7}.$ Thus, $x^2 = \frac{4}{49}$ , so the fraction of the triangle (area $6$ ) covered by the square is $\frac{2}{147}$ . The answer is then $\boxed{\dfrac{145}{147}}$ .

Problem 24

答案：D
Solution 1：By angle bisector theorem, $BG=\frac{5a}{6}$ . By similar triangles, $DF=\frac{5a}{12}$ , and the height of this trapezoid is $\frac{h}{2}$ , where $h$ is the length of the altitude to $BC$ . Then $\frac{ah}{2}=120$ and we wish to compute $\frac{5a}{8}\cdot\frac{h}{2}=\boxed{75}$ .
Solution 2： $\overline{DE}$ is midway from $A$ to $\overline{BC}$ , and $DE = \frac{BC}{2}$ . Therefore, $\bigtriangleup ADE$ is a quarter of the area of $\bigtriangleup ABC$ , which is $30$ . Subsequently, we can compute the area of quadrilateral $BDEC$ to be $120 - 90 = 30$ . Using the angle bisector theorem in the same fashion as the previous problem, we get that $\overline{BG}$ is $5$ times the length of $\overline{GC}$ . We want the larger piece, as described by the problem. Because the heights are identical, one area is $5$ times the other, and $\frac{5}{6} \cdot 90 = \boxed{75}$ .

Problem 25

答案：D
Solution：Observe $A_n = a(1 + 10 + \dots + 10^{n - 1}) = a \cdot \tfrac{10^n - 1}{9}$ ; similarly $B_n = b \cdot \tfrac{10^n - 1}{9}$ and $C_n = c \cdot \tfrac{10^{2n} - 1}{9}$ . The relation $C_n - B_n = A_n^2$ rewrites as $c \cdot \frac{10^{2n} - 1}{9} - b \cdot \frac{10^n - 1}{9} = a^2 \cdot \left(\frac{10^n - 1}{9}\right)^2.$ Since $n > 0$ , $10^n > 1$ and we may cancel out a factor of $\tfrac{10^n - 1}{9}$ to obtain $c \cdot (10^n + 1) - b = a^2 \cdot \frac{10^n - 1}{9}.$ This is a linear equation in $10^n$ . Thus, if two distinct values of $n$ satisfy it, then all values of $n$ will. Matching coefficients, we need $c = \frac{a^2}{9} \quad \text{and} \quad c - b = -\frac{a^2}{9} \implies b = \frac{2a^2}{9}.$ To maximize $a + b + c = a + \tfrac{a^2}{3}$ , we need to maximize $a$ . Since $b$ and $c$ must be integers, $a$ must be a multiple of 3. If $a = 9$ then $b$ exceeds 9. However, if $a = 6$ then $b = 8$ and $c = 4$ for an answer of $\boxed{\textbf{(D)} \text{ 18}}$ . (CantonMathGuy)