2025 AIME II Problems/Problem 1
Problem
Six points 
and 
lie in a straight line in that order. Suppose that 
is a point not on the line and that 
and 
Find the area of 
Solution 1
A=(0,0);
label("$A$", A, S);
B=(1.5,0);
label("$B$", B, S);
C=(2.9,0);
label("$C$", C, S);
D=(4.2,0);
label("$D$", D, S);
E=(5.3,0);
label("$E$", E, S);
F=(6.5,0);
label("$F$", F, S);
G=(3.7,3);
label("$G$", G, N);
draw(A--B--C--D--E--F);
draw(C--G--D);
draw(B--G--E);
(Error making remote request. Unknown error_msg)
Let 
, 
, 
, 
and 
. Then we know that 
=73, 
, 
, 
and 
. From this we can easily deduce 
and 
thus 
. Using Heron's formula we can calculate the area of 
to be 
, and since the base of 
is 
of that of , we calculate the area of to be 
.
Solution 2 (Law of Cosines)
We need to solve the lengths of AB, BC, CD, DE and EF. Let AB = a, BC = b, CD = c, DE = d and EF = e a + b = 26, b + c = 22, c + d = 31, d + e = 33, a + b + c + d + e = 73 Substituting a + b = 26 and d + e = 33 into a + b + c + d + e = 73, c = 14 Therefore, we get a = 18, b = 8, c = 14, d = 17 and e = 16 Noting that triangle CDG has CD = 14, CG = 40 and DG = 30, 14^2 = 40^2 + 30^2 - 2 * 40 * 30 * cos(angle CGD) 2 * 40 * 30 * cos(angle CGD) = 2500 - 196 = 2304 cos(angle CGD) = 24/25, therefore sin(angle CGD) The area for triangle CDG = 1/2 * 40 * 30 * 7/25 = 168 Noting that the height of the triangle CDG would be the exact same height as the triangle as triangle BGE, the ratio of the length would be the ratio of area. Therefore, the answer would be given by 168 * 39 / 14 = 468
(Feel free to correct any latex and formats) ~Mitsuihisashi14
See also
| 2025 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
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