2018 AMC 10A Solutions | LIVE by Po-Shen Loh

1.

What is the value of $\left(\left((2+1)^{-1}+1\right)^{-1}+1\right)^{-1}+1?$

$\dfrac{5}{8}$

$\dfrac{11}{7}$

$\dfrac{8}{5}$

$\dfrac{18}{11}$

$\dfrac{15}{8}$

Concepts:fraction order of operations

Difficulty rating: 560

Small Hint:

Work from the innermost reciprocal outward

Big Hint:

After each reciprocal step, simplify before adding the next $1$

Solution:

We can simplify this as follows. $\begin{aligned} &\left(\left((2+1)^{-1}+1\right)^{-1}+1\right)^{-1}+1 \\ &=\left(\left(\dfrac{1}{3}+1\right)^{-1}+1\right)^{-1}+1 \\ &=\left(\dfrac{3}{4}+1\right)^{-1}+1 \\ &=\dfrac{4}{7} + 1 \\ &=\dfrac{11}{7} \end{aligned}$

Thus, B is the correct answer.

2.

Liliane has $50\%$ more soda than Jacqueline, and Alice has $25\%$ more soda than Jacqueline. What is the relationship between the amounts of soda that Liliane and Alice have?

Liliane has $20\%$ more soda than Alice.

Liliane has $25\%$ more soda than Alice.

Liliane has $45\%$ more soda than Alice.

Liliane has $75\%$ more soda than Alice.

Liliane has $100\%$ more soda than Alice.

Concepts:percentage ratio and proportion

Difficulty rating: 770

Small Hint:

Use Jacqueline’s amount as the base variable

Big Hint:

Compare Liliane’s amount directly to Alice’s amount

Solution:

Let $x$ be the number of gallons of soda that Jacqueline has. Then Alice has $1.25x$ gallons, and Liliane has $1.5x$ gallons.

Therefore, the relationship can be found by dividing the amount of soda that each has to yield $\dfrac{1.5x}{1.25x}=1.2,$ which means Liliane has $20\%$ more soda than Alice.

Thus, A is the correct answer.

3.

A unit of blood expires after $10! = 10 \cdot 9 \cdot 8 \cdots 1$ seconds. Yasin donates a unit of blood at noon on January $1.$ On what day does his unit of blood expire?

January $2$

January $12$

January $22$

February $11$

February $12$

Concepts:factorial unit conversion date and time

Difficulty rating: 960

Small Hint:

Convert $10!$ seconds into days

Big Hint:

Use $60\cdot60\cdot24$ seconds per day

Solution:

We can divide $10!$ by $60,$ $60,$ and $24$ to get the number of days that it takes for a unit of blood to expire.

The first division cancels a $6$ and $10.$ The second division cancels $3, 4,$ and $5.$ The final division cancels $8$ and turns the $9$ into a $3.$

This leaves $2, 7,$ and a $3,$ which multiply to $42.$ There are $31$ days in January, so by February $1,$ the blood only has $42 - 31 = 11$ days left.

$11$ days from February $1$ would make the blood expire on February $12.$

Thus, E is the correct answer.

4.

How many ways can a student schedule $3$ mathematics courses — algebra, geometry, and number theory — in a $6$ -period day if no two mathematics courses can be taken in consecutive periods?

(What courses the student takes during the other $3$ periods is of no concern here.)

$3$

$6$

$12$

$18$

$24$

Concepts:arrangements with restrictions permutations systematic listing

Difficulty rating: 1220

Small Hint:

First choose the three nonconsecutive periods

Big Hint:

After the periods are chosen, arrange the three named courses

Solution:

The $3$ classes can occupy the following periods: $(1, 3, 5),$ $(1, 3, 6),$ $(1, 4, 6),$ $(2, 4, 6).$

This means that there are $4$ ways to choose which periods the mathematics courses occur.

For each configuration, there are $3!$ ways to determine the order of the courses, for a total of $6 \cdot 4 = 24$ schedules.

Thus, E is the correct answer.

5.

Alice, Bob, and Charlie were on a hike and were wondering how far away the nearest town was. When Alice said, “We are at least $6$ miles away,” Bob replied, “We are at most $5$ miles away.” Charlie then remarked, “Actually the nearest town is at most $4$ miles away.” It turned out that none of the three statements was true. Let $d$ be the distance in miles to the nearest town. Which of the following intervals is the set of all possible values of $d?$

$(0,4)$

$(4,5)$

$(4,6)$

$(5,6)$

$(5,\infty)$

Concepts:logical deduction inequality

Difficulty rating: 900

Small Hint:

Negate each person’s statement

Big Hint:

Intersect the three resulting inequalities for $d$

Solution:

Alice’s statement tells us that $d \lt 6.$ Bob’s statement tells us that $d \gt 5.$ Charlie’s statement tells us that $d \gt 4.$

Combining all of these tells us that $5 \lt d$ and $d \lt 6,$ which means $d$ is in the interval $(5, 6).$

Thus, D is the correct answer.

6.

Sangho uploaded a video to a website where viewers can vote that they like or dislike a video. Each video begins with a score of $0,$ and the score increases by $1$ for each like vote and decreases by $1$ for each dislike vote.

At one point Sangho saw that his video had a score of $90,$ and that $65\%$ of the votes cast on his video were like votes. How many votes had been cast on Sangho’s video at that point?

$200$

$300$

$400$

$500$

$600$

Concepts:percentage linear equation

Difficulty rating: 900

Small Hint:

Let $N$ be the total number of votes

Big Hint:

The score is likes minus dislikes

Solution:

If $65\%$ of votes were like votes, then $35\%$ of votes are dislike votes. Then Sangho’s score is $65\% - 35\% = 30\%$ the total number of votes.

We know that Sangho’s score is $90,$ so the total number of votes is $90 \div 30\% = 300.$

Thus, B is the correct answer.

7.

For how many (not necessarily positive) integer values of $n$ is the following value an integer? $4000 \cdot \left(\dfrac{2}{5}\right)^n$

$3$

$4$

$6$

$8$

$9$

Concepts:exponent prime factorization counting integers in a range

Difficulty rating: 1070

Small Hint:

Write $4000$ as $2^5\cdot5^3$

Big Hint:

The exponents of $2$ and $5$ must both be nonnegative

Solution:

We can rewrite the expression as $(2^5 \cdot 5^3) \cdot \left(\dfrac{2}{5}\right)^n = 2^{5 + n} \cdot 5^{3 - n}.$

For this to be an integer, both exponents must be nonnegative. This means that $\begin{aligned} 5 + n \geq 0 &\Rightarrow n \geq -5 \\ 3 - n \geq 0 &\Rightarrow n \leq 3. \end{aligned}$

This gives us $5 + 3 + 1 = 9$ values for $n.$

Thus, E is the correct answer.

8.

Joe has a collection of $23$ coins, consisting of $5$ -cent coins, $10$ -cent coins, and $25$ -cent coins. He has $3$ more $10$ -cent coins than $5$ -cent coins, and the total value of his collection is $320$ cents. How many more $25$ -cent coins does Joe have than $5$ -cent coins?

$0$

$1$

$2$

$3$

$4$

Concepts:money system of equations

Difficulty rating: 1220

Small Hint:

Let the number of $5$ -cent coins be $x$

Big Hint:

Use both the total number of coins and total value equations

Solution:

Let $x$ be the number of $5$ -cent coins that Joe has. Then the number of $10$ -cent coins he has is $x + 3.$

Therefore, Joe has $23 - x - (x + 3) = 20 - 2x$ $25$ -cent coins.

The total value of all these coins is $5x + 10(x + 3) + 25(20 - 2x)$ $= 530 - 35x.$

We know that $530 - 35x = 320 \Rightarrow x = 6.$

This means that Joe has $20 - 2 \cdot 6 = 8$ $25$ -cent coins. Therefore, he has $8 - 6 = 2$ more $25$ -cent coins than $5$ -cent coins.

Thus, C is the correct answer.

9.

All of the triangles in the diagram below are similar to isosceles triangle $ABC,$ in which $AB=AC.$ Each of the $7$ smallest triangles has area $1,$ and $\triangle ABC$ has area $40.$ What is the area of trapezoid $DBCE?$

$16$

$18$

$20$

$22$

$24$

Concepts:similarity area ratio

Difficulty rating: 1420

Small Hint:

Area scales as the square of side length for similar triangles

Big Hint:

Find the area of the top triangle and subtract from $40$

Solution:

We know that the side length of the smaller triangles is $\sqrt{\frac{1}{40}}$ times the length of the larger triangle from similar triangles.

Then the side length of $\triangle ADE$ is $4\sqrt{\frac{1}{40}}$ times the length of the side length of the larger triangle.

This makes the ratio of the areas $\left(4\sqrt{\dfrac{1}{40}}\right)^2 = 16 \cdot \dfrac{1}{40} = \dfrac{2}{5}.$

Therefore, the area of $\triangle ADE$ is $\frac{2}{5} \cdot 40 = 16.$ The area of the trapezoid is then $40 - 16 = 24.$

Thus, E is the correct answer.

10.

Suppose that real number $x$ satisfies $\sqrt{49-x^2}-\sqrt{25-x^2}=3.$ What is the value of $\sqrt{49-x^2}+\sqrt{25-x^2}?$

$8$

$\sqrt{33}+3$

$9$

$2\sqrt{10}+4$

$12$

Concepts:radical difference of squares algebraic manipulation

Difficulty rating: 1310

Small Hint:

Treat the two radicals as conjugate-like quantities

Big Hint:

Multiply the given difference by the desired sum

Solution:

Note that the left hand side of the equation and the desired expression are conjugates. Multiplying them would remove the square roots.

Multiplying them yields $49 - x^2 - 25 + x^2 = 24.$

This means that the product of the values of the expressions is equal to $24.$ The desired value is therefore $24 \div 3 = 8.$

Thus, A is the correct answer.

11.

When $7$ fair standard $6$ -sided dice are thrown, the probability that the sum of the numbers on the top faces is $10$ can be written as $\dfrac{n}{6^{7}},$ where $n$ is a positive integer. What is $n?$

$42$

$49$

$56$

$63$

$84$

Concepts:dice (probability)stars and bars

Difficulty rating: 1420

Small Hint:

List the unordered ways seven positive die rolls can sum to $10$

Big Hint:

Count the orderings for each listed pattern

Solution:

We can use stars and bars to find $n.$ It is the same as finding the number of ways to put $10$ balls into $7$ boxes, where each box has at least one ball.

The formula for such a scenario is $\binom{n - 1}{k - 1},$ where $n$ is the number of balls and $k$ is the number of boxes.

No die can exceed $4$ in a sum of $10$ from seven positive rolls, so the upper bound of $6$ creates no additional restriction. The desired answer is therefore $\binom{9}{6} = \binom{9}{3} = 84.$

Thus, E is the correct answer.

12.

How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations? $\begin{cases} ~x+3y&=3 \\ ~\big||x|-|y|\big|&=1 \end{cases}$

$1$

$2$

$3$

$4$

$8$

Concepts:absolute value system of equations casework

Difficulty rating: 1370

Small Hint:

Replace the absolute-value equation by four linear cases

Big Hint:

Distinct solutions can repeat across cases, so count unique ordered pairs

Solution:

The second equation says $|x|-|y|=1$ or $|x|-|y|=-1$ , so it is enough to check the four linear possibilities $x=y\pm1$ and $x=-y\pm1$ .

Combining these with $x+3y=3$ gives $(x,y)=\left(\dfrac32,\dfrac12\right)$ , $(0,1)$ , $(0,1)$ again, and $(-3,2)$ .

These are three distinct ordered pairs, and each satisfies the original absolute-value equation. Thus, C is the correct answer.

13.

A paper triangle with sides of lengths $3,$ $4,$ and $5$ inches, as shown, is folded so that point $A$ falls on point $B.$ What is the length in inches of the crease?

$1+\dfrac{1}{2} \sqrt{2}$

$\sqrt{3}$

$\dfrac{7}{4}$

$\dfrac{15}{8}$

$2$

Concepts:paper folding perpendicular bisector similarity

Difficulty rating: 1420

Small Hint:

The crease is the perpendicular bisector of $AB$

Big Hint:

Use similarity with the original $3$ - $4$ - $5$ triangle

Solution:

The crease is the perpendicular bisector of $\overline{AB}.$ Let $\overline{DE}$ be the crease.

By $AA$ similarity, $\triangle ADE\sim\triangle ACB.$ Therefore, $\dfrac{BC}{AC}=\dfrac{DE}{AD}.$ Plugging in the side lengths gives $\dfrac34=\dfrac{DE}{\frac{5}{2}},$ so $DE=\dfrac{15}{8}.$

Thus, D is the correct answer.

14.

What is the greatest integer less than or equal to $\dfrac{3^{100}+2^{100}}{3^{96}+2^{96}}?$

$80$

$81$

$96$

$97$

$625$

Concepts:exponent floor and ceiling functions bounding to limit cases

Difficulty rating: 1540

Small Hint:

Compare the expression to $81$ first

Big Hint:

Then prove it is still greater than $80$

Solution:

Let $a=3^{96}$ and $b=2^{96}$ . The expression is $\dfrac{81a+16b}{a+b}=16+\dfrac{65a}{a+b}$ , so it is less than $16+65=81$ .

To show the floor is $80$ , we also need the expression to be greater than $80$ . This is equivalent to $81a+16b>80a+80b$ , or $a>64b$ .

Because $\left(\dfrac32\right)^2=\dfrac94>2,$ we have $\dfrac{a}{b}=\left(\dfrac32\right)^{96}>2^{48}>64.$ Hence the expression is greater than $80$ and less than $81.$ Thus, A is the correct answer.

15.

Two circles of radius $5$ are externally tangent to each other and are internally tangent to a circle of radius $13$ at points $A$ and $B,$ as shown in the diagram. The distance $AB$ can be written in the form $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. What is $m+n?$

$21$

$29$

$58$

$69$

$93$

Concepts:tangent circles similarity

Difficulty rating: 1820

Small Hint:

Join the three circle centers

Big Hint:

The triangle through tangent points is similar to the triangle through centers

Solution:

Let $X$ be the center of the large circle and let $Y,Z$ be the centers of the two smaller circles. Then $XY=XZ=13-5=8$ and $YZ=10$ .

The radii to tangent points make $XA$ collinear with $XY$ and $XB$ collinear with $XZ$ , so $\triangle XAB\sim \triangle XYZ$ . Thus $\dfrac{AB}{YZ}=\dfrac{XA}{XY}=\dfrac{13}{8}$ .

Hence $AB=10\cdot\dfrac{13}{8}=\dfrac{65}{4}$ , so $m+n=65+4=69$ . Thus, D is the correct answer.

16.

Right triangle $ABC$ has leg lengths $AB=20$ and $BC=21.$ Including $\overline{AB}$ and $\overline{BC},$ how many line segments with integer length can be drawn from vertex $B$ to a point on hypotenuse $\overline{AC}?$

$5$

$8$

$12$

$13$

$15$

Concepts:right triangle altitude counting integers in a range

Difficulty rating: 1540

Small Hint:

Find the altitude from $B$ to the hypotenuse

Big Hint:

Integer-length segments occur in symmetric pairs around the altitude

Solution:

Let $P$ be the foot of the altitude from $B$ to $\overline{AC}.$ The Pythagorean Theorem gives $AC=29.$ Computing the area in two ways gives $\dfrac{29\cdot PB}{2}=\dfrac{20\cdot21}{2},$ so $PB=\dfrac{420}{29},$ which lies between $14$ and $15.$

As the endpoint moves from $A$ to $P,$ its distance from $B$ decreases continuously from $20$ to $PB.$ Thus there is one segment of each integer length $15,16,17,18,19,20.$ As the endpoint moves from $P$ to $C,$ the distance increases continuously from $PB$ to $21,$ giving one segment of each integer length $15,16,17,18,19,20,21.$ These are $6+7=13$ distinct segments.

Thus, D is the correct answer.

17.

Let $S$ be a set of $6$ integers taken from $\{1,2,\dots,12\}$ with the property that if $a$ and $b$ are elements of $S$ with $a < b,$ then $b$ is not a multiple of $a.$ What is the least possible value of an element in $S?$

$2$

$3$

$4$

$5$

$7$

Concepts:divisibility extremal argument casework

Difficulty rating: 1970

Small Hint:

Group numbers so at most one from each group can be chosen

Big Hint:

Try forcing six choices from the divisor chains

Solution:

We proceed by casing on possible values for $S:$

$1$ cannot be the smallest element since that would mean that no other number can be in the set.

$2$ cannot be the smallest element since we would have to include every odd number except $1.$ This would make $3$ and $9$ violate the rule.

Let $3$ be the smallest element. Then we can include $7$ and $11.$ We can finally include either $4$ or $8$ and $5$ or $10.$

Either way, the maximum number of elements that we can include is $5,$ so $3$ cannot be the smallest element.

Starting with $4,$ we can include $6, 7, 9$ and $11.$ Finally, we can add either $5$ or $10,$ creating a $6$ -element set.

Thus, C is the correct answer.

18.

How many nonnegative integers can be written in the form $a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5$ $+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2$ $+a_1\cdot3^1+a_0\cdot3^0,$ where $a_i\in \{-1,0,1\}$ for $0\le i \le 7?$

$512$

$729$

$1094$

$3281$

$59{,}048$

Concepts:number base symmetry pairing and grouping

Difficulty rating: 1770

Small Hint:

Think of the expression as balanced ternary

Big Hint:

Positive and negative values pair off symmetrically around $0$

Solution:

Note that every number formed by this sum is either positive, negative, or zero.

The number of positive numbers equals the number of negative numbers due to symmetry (flip the $1$ s to $-1$ s and $-1$ s to $1$ s).

The only way for the sum to be $0$ is if all the coefficients are $0.$

The total number of numbers is $3^8 = 6561.$ Because each power of $3$ is larger than the sum of all previous powers of three, each combination of coefficients yields a different value. More explicitly, at the highest place where two combinations differ, the difference has magnitude at least $3^k,$ while all lower places together can cancel at most $2(1+3+\cdots+3^{k-1})=3^k-1.$

Therefore, there are $\dfrac{6561 - 1}{2} + 1 = 3281$ distinct nonnegative integers.

Thus, D is the correct answer.

19.

A number $m$ is randomly selected from the set $\{11,13,15,17,19\},$ and a number $n$ is randomly selected from $\{1999,2000,2001,\ldots,2018\}.$ What is the probability that $m^n$ has a units digit of $1?$

$\dfrac{1}{5}$

$\dfrac{1}{4}$

$\dfrac{3}{10}$

$\dfrac{7}{20}$

$\dfrac{2}{5}$

Concepts:units digit modular exponentiation basic probability casework

Difficulty rating: 1540

Small Hint:

Only units-digit cycles matter

Big Hint:

Average the success counts for bases ending in $1,3,5,7,9$

Solution:

Only the units digit of $m$ matters. Among the $20$ consecutive possible exponents, each residue modulo $4$ occurs $5$ times, and $10$ exponents are even. A base ending in $1$ succeeds for all $20$ exponents; bases ending in $3$ or $7$ succeed for the $5$ exponents divisible by $4;$ a base ending in $5$ never succeeds; and a base ending in $9$ succeeds for the $10$ even exponents.

Thus $20+5+0+5+10=40$ of the $5\cdot20=100$ equally likely pairs work, and the probability is $\dfrac{40}{100}=\dfrac25.$

Thus, E is the correct answer.

20.

A scanning code consists of a $7 \times 7$ grid of squares, with some of its squares colored black and the rest colored white. There must be at least one square of each color in this grid of $49$ squares.

A scanning code is called symmetric if its look does not change when the entire square is rotated by a multiple of $90^{\circ}$ counterclockwise around its center, nor when it is reflected across a line joining opposite corners or a line joining midpoints of opposite sides.

What is the total number of possible symmetric scanning codes?

$510$

$1022$

$8190$

$8192$

$65{,}534$

Concepts:symmetry multiplication principle

Difficulty rating: 1970

Small Hint:

Classify grid squares by symmetry orbits

Big Hint:

Choose one color per orbit, excluding the two constant colorings

Solution:

Number rows and columns from $-3$ through $3,$ with the center at $(0,0).$ Rotations and reflections can change signs and exchange coordinates, so the orbit of $(x,y)$ is determined by the ordered pair obtained by sorting $(|x|,|y|).$ These pairs are $(u,v)$ with $0\le u\le v\le3,$ of which there are $4+3+2+1=10.$

Once one square in each orbit is colored, symmetry forces the colors of all other squares in that orbit.

There are therefore $2^{10}$ symmetric colorings before the condition about using both colors. The all-black and all-white colorings are not allowed.

The total number of valid symmetric scanning codes is $2^{10}-2=1022$ . Thus, B is the correct answer.

21.

Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$ -plane intersect at exactly $3$ points?

$a = \dfrac14$

$\dfrac14 \lt a \lt \dfrac12$

$a \gt \dfrac14$

$a = \dfrac12$

$a \gt \dfrac12$

Concepts:parabola circle substitution

Difficulty rating: 1820

Small Hint:

Substitute the parabola equation into the circle equation

Big Hint:

The nonzero roots appear only when $2a-1$ is positive

Solution:

Substitute $y=x^2-a$ into $x^2+y^2=a^2$ . This gives $x^2+(x^2-a)^2=a^2$ , so $x^2(x^2-(2a-1))=0$ .

The factor $x^2=0$ always gives the single point $(0,-a)$ . The other factor gives two additional real points exactly when $2a-1>0$ .

There are exactly three intersection points when $a>\dfrac12$ . Thus, E is the correct answer.

22.

Let $a,$ $b,$ $c,$ and $d$ be positive integers such that $\gcd(a, b)=24,$ $\gcd(b, c)=36,$ $\gcd(c, d)=54,$ and $70 < \gcd(d, a) < 100.$ Which of the following must be a divisor of $a?$

$5$

$7$

$11$

$13$

$17$

Concepts:greatest common divisor prime factorization bounding to limit cases

Difficulty rating: 2010

Small Hint:

Track the forced powers of $2$ and $3$ in the gcds

Big Hint:

The remaining factor of $\gcd(d,a)$ must lie between $12$ and $17$

Solution:

From $\gcd(a,b)=24=2^3\cdot3$ and $\gcd(b,c)=36=2^2\cdot3^2$ , the number $a$ is divisible by $2^3\cdot3$ but not by $3^2$ .

From $\gcd(b,c)=36$ and $\gcd(c,d)=54=2\cdot3^3$ , the number $d$ is divisible by $2\cdot3^3$ but not by $2^2$ . Therefore $\gcd(d,a)=2\cdot3\cdot n$ , where $n$ has no factor $2$ or $3$ .

Since $70<6n<100,$ the integer $n$ satisfies $12\le n\le16.$ The only value in this range with no factor $2$ or $3$ is $13,$ so $13$ must divide $a.$ Thus, D is the correct answer.

23.

Farmer Pythagoras has a field in the shape of a right triangle. The right triangle’s legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from $S$ to the hypotenuse is $2$ units. What fraction of the field is planted?

$\dfrac{25}{27}$

$\dfrac{26}{27}$

$\dfrac{73}{75}$

$\dfrac{145}{147}$

$\dfrac{74}{75}$

Concepts:area decomposition right triangle

Difficulty rating: 2150

Small Hint:

Let $x$ be the side length of the unplanted square

Big Hint:

Decompose the triangle into the square, two right triangles, and a triangle of height $2$

Solution:

Let $x$ be the side length of $S.$ Then we can split the field up into the following shapes.

We can express the area of the field in two ways: $\dfrac{3 \cdot 4}{2} = x^2 + \dfrac{x(3 - x)}{2}$ $+ \dfrac{x(4 - x)}{2} + \dfrac{2 \cdot 5}{2}.$

Simplifying yields $6 = \dfrac{7x}{2} + 5$ $x = \dfrac{2}{7}.$

The desired fraction is $\dfrac{6 - x^2}{6} = \dfrac{6 - \frac{4}{49}}{6} = \dfrac{145}{147}.$

Thus, D is the correct answer.

24.

Triangle $ABC$ with $AB=50$ and $AC=10$ has area $120.$ Let $D$ be the midpoint of $\overline{AB},$ and let $E$ be the midpoint of $\overline{AC}.$ The angle bisector of $\angle BAC$ intersects $\overline{DE}$ and $\overline{BC}$ at $F$ and $G,$ respectively. What is the area of quadrilateral $FDBG?$

$60$

$65$

$70$

$75$

$80$

Concepts:angle bisector theorem midpoint trapezoid area ratio

Difficulty rating: 2040

Small Hint:

Use the angle bisector theorem to locate $G$

Big Hint:

The midsegment $DE$ makes a similar half-height trapezoid

Solution:

Let $BC = a, BG = x, GC = y,$ and $h$ be the length of the altitude through $A.$

By the Angle Bisector Theorem, $BG:GC=AB:AC=5:1,$ so $BG=\dfrac{5a}{6}.$ Because $DE$ is a midsegment, $DE=\dfrac a2.$ Applying the same theorem in $\triangle ADE$ gives $DF:FE=AD:AE=5:1,$ so $DF=\dfrac56DE=\dfrac{5a}{12}.$

The trapezoid’s height is $\frac{h}{2},$ and $\frac{ah}{2}=120.$ Its average base length is $\dfrac12\left(\dfrac{5a}{12}+\dfrac{5a}{6}\right)=\dfrac{5a}{8}.$ Therefore, its area is $\dfrac{5a}{8}\cdot\dfrac h2=\dfrac58\cdot\dfrac{ah}{2}=75.$ Thus, D is the correct answer.

25.

For a positive integer $n$ and nonzero digits $a,$ $b,$ and $c,$ let $A_n$ be the $n$ -digit integer each of whose digits is equal to $a$ ; let $B_n$ be the $n$ -digit integer each of whose digits is equal to $b$ ; and let $C_n$ be the $2n$ -digit (not $n$ -digit) integer each of whose digits is equal to $c.$ What is the greatest possible value of $a + b + c$ for which there are at least two values of $n$ such that $C_n - B_n = A_n^2?$

$12$

$14$

$16$

$18$

$20$

Concepts:digits geometric sequence algebraic manipulation

Difficulty rating: 2390

Small Hint:

Write the repeated-digit numbers as geometric sums

Big Hint:

Having two different $n$ values forces the linear equation in $10^n$ to be identically true

Solution:

The repeated-digit numbers can be written as $\begin{aligned} A_n&=a\dfrac{10^n-1}{9},\\ B_n&=b\dfrac{10^n-1}{9},\\ C_n&=c\dfrac{10^{2n}-1}{9}. \end{aligned}$

Substituting these expressions into $C_n-B_n=A_n^2$ and dividing by $\frac{10^n-1}{9}$ gives $c(10^n+1)-b=a^2\dfrac{10^n-1}{9}.$

Rearranging yields $(9c-a^2)10^n=9b-9c-a^2.$ If this holds for two different values of $n,$ subtracting the two equations shows that $9c-a^2=0.$ The displayed equation then also forces $9b-9c-a^2=0.$

Hence $c=\frac{a^2}{9}$ and $b=\frac{2a^2}{9}.$ Because $a,b,c$ are nonzero digits, the candidates are $(a,b,c)=(3,2,1),$ $(6,8,4),$ and $(9,18,9).$ The last triple is invalid because $18$ is not a digit. The greatest valid sum is $6+8+4=18.$ Thus, D is the correct answer.