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

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jopok is 2=2
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==Problem==
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On <math>\triangle ABC</math> points <math>A</math>, <math>D</math>, <math>E</math>, and <math>B</math> lie in that order on side <math>\overline{AB}</math> with <math>AD = 4</math>, <math>DE = 16</math>, and <math>EB = 8</math>. Points <math>A</math>, <math>F</math>, <math>G</math>, and <math>C</math> lie in that order on side <math>\overline{AC}</math> with <math>AF = 13</math>, <math>FG = 52</math>, and <math>GC = 26</math>. Let <math>M</math> be the reflection of <math>D</math> through <math>F</math>, and let <math>N</math> be the reflection of <math>G</math> through <math>E</math>. Quadrilateral <math>DEGF</math> has area <math>288</math>. Find the area of heptagon <math>AFNBCEM</math>.
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<asy>
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unitsize(14);
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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;
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filldraw(A--F--N--B--C--E--M--cycle, lightgray);
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draw(A--B--C--cycle);
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draw(D--M);
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draw(N--G);
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dot(A);
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dot(B);
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dot(C);
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dot(D);
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dot(E);
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dot(F);
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dot(G);
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dot(M);
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dot(N);
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label("$A$", A, dir(90));
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label("$B$", B, dir(225));
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label("$C$", C, dir(315));
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label("$D$", D, dir(135));
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label("$E$", E, dir(135));
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label("$F$", F, dir(45));
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label("$G$", G, dir(45));
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label("$M$", M, dir(45));
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label("$N$", N, dir(135));
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</asy>
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 +
==Video solution by [[User:grogg007|grogg007]]==
 +
https://youtu.be/PNBxBvvjbcU?t=80
 +
 
 +
==Solution 1==
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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>.
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 +
~alwaysgonnagiveyouup
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==Solution 2==
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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>, are similar, 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 <math>12</math>, and finally multiplying <math>12 \cdot 49</math> gives us our answer of <math>\boxed{588}</math>
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~shreyan.chethan, cleaned up by cweu001
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==Solution 3==
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By area lemma, we can see that the areas of the shaded areas are equivalent to the areas of the unshaded areas. Thus, we see that the desired area is equivalent to the area of the triangle <math>\triangle ABC</math>. Since <math>AF : FG : GC = 1 : 4 : 2</math>, we have <math>[ADF]:[AEG]:[ABC] = 1:25:49</math>, meaning <math>[ADF]:[DEGF]:[BEGC] = 1:24:24</math>. Thus, since <math>\frac{[DEGF]}{ABC} = \frac{24}{49}</math>, we can calculate <math>[ABC] = 588</math>.
 +
 
 +
~cweu001, cleaned up by 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==
 +
 
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https://www.youtube.com/watch?v=fVBk2vOusio&t=3s
 +
 
 +
~GP102
 +
 
 +
==Video Solution by yjtest==
 +
 
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https://www.youtube.com/watch?v=avaHHEOQEZs
 +
 
 +
==See also==
 +
{{AIME box|year=2025|num-b=1|num-a=3|n=I}}
 +
 
 +
{{MAA Notice}}

Latest revision as of 21:51, 16 July 2025

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]

Video solution by grogg007

https://youtu.be/PNBxBvvjbcU?t=80

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, cleaned up by cweu001

Solution 3

By area lemma, we can see that the areas of the shaded areas are equivalent to the areas of the unshaded areas. Thus, we see that the desired area is equivalent to the area of the triangle $\triangle ABC$. Since $AF : FG : GC = 1 : 4 : 2$, we have $[ADF]:[AEG]:[ABC] = 1:25:49$, meaning $[ADF]:[DEGF]:[BEGC] = 1:24:24$. Thus, since $\frac{[DEGF]}{ABC} = \frac{24}{49}$, we can calculate $[ABC] = 588$.

~cweu001, cleaned up by 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

Video Solution by yjtest

https://www.youtube.com/watch?v=avaHHEOQEZs

See also

2025 AIME I (Problems • Answer Key • Resources)
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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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