2025 AIME I Problems/Problem 2

Problem

On $\triangle ABC$ points $A$ , $D$ , $E$ , and $B$ lie in that order on side $\overline{AB}$ with $AD = 4$ , $DE = 16$ , and $EB = 8$ . Points $A$ , $F$ , $G$ , and $C$ lie in that order on side $\overline{AC}$ with $AF = 13$ , $FG = 52$ , and $GC = 26$ . Let $M$ be the reflection of $D$ through $F$ , and let $N$ be the reflection of $G$ through $E$ . Quadrilateral $DEGF$ has area $288$ . Find the area of heptagon $AFNBCEM$ .

$[asy] unitsize(14); pair A = (0, 9), B = (-6, 0), C = (12, 0), D = (5A + 2B)/7, E = (2A + 5B)/7, F = (5A + 2C)/7, G = (2A + 5C)/7, M = 2F - D, N = 2E - G; filldraw(A--F--N--B--C--E--M--cycle, lightgray); draw(A--B--C--cycle); draw(D--M); draw(N--G); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(G); dot(M); dot(N); label("$A$", A, dir(90)); label("$B$", B, dir(225)); label("$C$", C, dir(315)); label("$D$", D, dir(135)); label("$E$", E, dir(135)); label("$F$", F, dir(45)); label("$G$", G, dir(45)); label("$M$", M, dir(45)); label("$N$", N, dir(135)); [/asy]$

Solution 1

Note that the triangles outside $\triangle ABC$ have the same height as the unshaded triangles in $\triangle ABC$ . Since they have the same bases, the area of the heptagon is the same as the area of triangle $ABC$ . Therefore, we need to calculate the area of $\triangle ABC$ . Denote the length of $DF$ as $x$ and the altitude of $A$ to $DF$ as $h$ . Since $\triangle ADF \sim \triangle AEG$ , $EG = 5x$ and the altitude of $DFGE$ is $4h$ . The area $[DFGE] = \frac{5x + x}{2} \cdot 4h = 3x \cdot 4h = 12xh = 288 \implies xh = 24$ . The area of $\triangle ABC$ is equal to $\frac{1}{2} 7x \cdot 7h = \frac{1}{2} 49xh = \frac{1}{2} 49 \cdot 24 = \frac{1}{2} 1176 = \boxed{588}$ .

~alwaysgonnagiveyouup

Solution 2

Because of reflections, and various triangles having the same bases, we can conclude that $|AFNBCEM| = |ABC|$ . Through the given lengths of $4-16-8$ on the left and $13-52-26$ on the right, we conclude that the lines through $\triangle ABC$ are parallel, and the sides are in a $1:4:2$ ratio. Because these lines are parallel, we can see that $ADF,~AEG,~ABC$ , are similar, and from our earlier ratio, we can give the triangles side ratios of $1:5:7$ , or area ratios of $1:25:49$ . Quadrilateral $DEGF$ corresponds to the $|AEG|-|ADF|$ , which corresponds to the ratio $25-1=24$ . Dividing $288$ by $24$ , we get $12$ , and finally multiplying $12 \cdot 49$ gives us our answer of $\boxed{588}$

~shreyan.chethan

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=J-0BapU4Yuk

Video Solution(Fast! Easy!)

https://youtu.be/LQyncubz30U

~MC

Video Solution by Mathletes Corner

https://www.youtube.com/watch?v=fVBk2vOusio&t=3s

~GP102

2025 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search