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Difference between revisions of "2025 AIME I Problems/Problem 2"

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~ alwaysgonnagiveyouup
 
~ alwaysgonnagiveyouup
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==Video Solution 1 by SpreadTheMathLove==
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https://www.youtube.com/watch?v=J-0BapU4Yuk
  
 
==See also==
 
==See also==

Revision as of 00:55, 14 February 2025

Problem

In $\triangle ABC$ points $D$ and $E$ lie on $\overline{AB}$ so that $AD < AE < AB$, while points $F$ and $G$ lie on $\overline{AC}$ so that $AF < AG < AC$. Suppose $AD = 4$, $DE = 16$, $EB = 8$, $AF = 13$, $FG = 52$, and $GC = 26$. Let $M$ be the reflection of $D$ through $F$, and let $N$ be the reflection of $G$ through $E$. The area of quadrilateral $DEGF$ is $288$. Find the area of heptagon $AFNBCEM$, as shown in the figure below. [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

Video Solution 1 by SpreadTheMathLove

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

See also

2025 AIME I (Problems • Answer Key • Resources)
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All AIME Problems and Solutions

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