Difference between revisions of "2025 AIME I Problems/Problem 2"
m (→Solution 1) |
|||
| Line 29: | Line 29: | ||
==Solution 1== | ==Solution 1== | ||
| − | 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 = 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 | ~ alwaysgonnagiveyouup | ||
| − | |||
==See also== | ==See also== | ||
Revision as of 20:13, 13 February 2025
Problem
In 
points 
and 
lie on 
so that 
, while points 
and 
lie on 
so that 
. Suppose 
, 
, 
, 
, 
, and 
. Let 
be the reflection of through , and let 
be the reflection of through . The area of quadrilateral 
is 
. Find the area of heptagon 
, 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]](http://latex.artofproblemsolving.com/6/4/b/64b0d84a95e0388e37622b54a251d836088fd97c.png)
Solution 1
Note that the triangles outside have the same height as the unshaded triangles in . Since they have the same bases, the area of the heptagon is the same as the area of triangle 
. Therefore, we need to calculate the area of . Denote the length of 
as 
and the altitude of 
to as 
. Since 
, 
and the altitude of 
is 
. The area ![$[DFGE] = \frac{5x + x}{2} \cdot 4h = 3x \cdot 4h = 12xh = 288 \implies xh = 24$](http://latex.artofproblemsolving.com/e/e/0/ee06844651656c7d34bda87a3fdb8f440ce455ad.png)
. The area of is equal to 
.
~ alwaysgonnagiveyouup
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
