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Difference between revisions of "2025 AIME I Problems/Problem 2"
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Note that the triangles outside <math>\triangle ABC</math> have the same height as the unshaded triangles in <math>\triangle ABC</math>. Since they have the same bases, the area of the heptagon is the same as the area of triangle <math>ABC</math>. Therefore, we need to calculate the area of <math>\triangle ABC</math>. Denote the length of <math>DF</math> as <math>x</math> and the altitude of <math>A</math> to <math>DF</math> as <math>h</math>. Since <math>\triangle ADF \sim \triangle AEG</math>, <math>EG = 5x</math> and the altitude of <math>DFGE</math> is <math>4h</math>. The area <math>[DFGE] = \frac{5x + x}{2} \cdot 4h = 3x \cdot 4h = 12xh = 288 \implies xh = 24</math>. The area of <math>\triangle ABC</math> is equal to <math>\frac{1}{2} 7x \cdot 7h = \frac{1}{2} 49xh = \frac{1}{2} 49 \cdot 24 = \frac{1}{2} 1176 = \boxed{588}</math>.
 
Note that the triangles outside <math>\triangle ABC</math> have the same height as the unshaded triangles in <math>\triangle ABC</math>. Since they have the same bases, the area of the heptagon is the same as the area of triangle <math>ABC</math>. Therefore, we need to calculate the area of <math>\triangle ABC</math>. Denote the length of <math>DF</math> as <math>x</math> and the altitude of <math>A</math> to <math>DF</math> as <math>h</math>. Since <math>\triangle ADF \sim \triangle AEG</math>, <math>EG = 5x</math> and the altitude of <math>DFGE</math> is <math>4h</math>. The area <math>[DFGE] = \frac{5x + x}{2} \cdot 4h = 3x \cdot 4h = 12xh = 288 \implies xh = 24</math>. The area of <math>\triangle ABC</math> is equal to <math>\frac{1}{2} 7x \cdot 7h = \frac{1}{2} 49xh = \frac{1}{2} 49 \cdot 24 = \frac{1}{2} 1176 = \boxed{588}</math>.
  
−
~ alwaysgonnagiveyouup
+
==Solution 2==
 +
Because of reflections, and various triangles having the same bases, we can conclude that <math>|AFNBCEM| = |ABC|</math>. Through the given lengths of <math>4-16-8</math> on the left and <math>13-52-26</math> on the right, we conclude that the lines through <math>\triangle ABC</math> are parallel, and the sides are in a <math>1:4:2</math> ratio. Because these lines are parallel, we can see that <math>ADF~AEG~ABC</math>, and from our earlier ratio, we can give the triangles side ratios of <math>1:5:7</math>, or area ratios of <math>1:25:49</math>. Quadrilateral <math>DEGF</math> corresponds to the <math>|AEG| - |ADF|</math>, which corresponds to the ratio <math>25-1=24</math>. Dividing <math>288</math> by <math>24</math>, we get 12, and finally multiplying <math>12 \cdot 49</math> gives us our answer of <math>\boxed{588}</math>
 +
 
 +
~shreyan.chethan
  
 
==Video Solution 1 by SpreadTheMathLove==
 
==Video Solution 1 by SpreadTheMathLove==

Revision as of 20:02, 15 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 $\frac{1}{2} 7x \cdot 7h = \frac{1}{2} 49xh = \frac{1}{2} 49 \cdot 24 = \frac{1}{2} 1176 = \boxed{588}$.

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 $ADF~AEG~ABC$, 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

See also

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

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