2025 AIME II Problems/Problem 1

Revision as of 22:51, 13 February 2025 by Lisztepos (talk | contribs) (Solution 1)

Problem

Six points $A, B, C, D, E,$ and $F$ lie in a straight line in that order. Suppose that $G$ is a point not on the line and that $AC=26, BD=22, CE=31, DF=33, AF=73, CG=40,$ and $DG=30.$ Find the area of $\triangle BGE.$

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 $AB=a$, $BC=b$, $CD=c$, $DE=d$ and $EF=e$. Then we know that $a+b+c+d+e$=73, $a+b=26$, $b+c=22$, $c+d=31$ and $d+e=33$. From this we can easily deduce $c=14$ and $a+e=34$ thus $b+c+d=39$. Using Heron's formula we can calculate the area of $\triangle{CGD}$ to be $\sqrt{(42)(28)(12)(2)}=168$, and since the base of $\triangle{BGE}$ is $\frac{39}{14}$ of that of $\triangle{CGD}$, we calculate the area of $\triangle{BGE}$ to be $168\times \frac{39}{14}=\boxed{468}$.

See also

2025 AIME II (ProblemsAnswer KeyResources)
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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