2025 AIME II Problems/Problem 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

$[asy] pair A,B,C,D,E,F,G; A=(0,0); label("$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$", G, N); draw(A--B--C--D--E--F); draw(C--G--D); draw(B--G--E); [/asy]$

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 $168\times \frac{39}{14}=\boxed{468}$ .

~ Quick Asymptote Fix by eevee9406, edited by aoum

Solution 2 (Law of Cosines)

We need to solve for the lengths of $AB$ , $BC$ , $CD$ , $DE$ , and $EF$ . Let $AB = a$ , $BC = b$ , $CD = c$ , $DE = d$ , and $EF = e$ . We are given the following system of equations:

$a + b = 26, \quad b + c = 22, \quad c + d = 31, \quad d + e = 33, \quad a + b + c + d + e = 73.$

Substituting $a + b = 26$ and $d + e = 33$ into the equation $a + b + c + d + e = 73$ , we get:

$c = 14.$

Thus, we have:

$a = 18, \quad b = 8, \quad c = 14, \quad d = 17, \quad e = 16.$

Next, consider triangle $CDG$ , where $CD = 14$ , $CG = 40$ , and $DG = 30$ . By the Law of Cosines, we have:

$CD^2 = CG^2 + DG^2 - 2 \times CG \times DG \times \cos(\angle CGD).$

Substituting the known values:

$14^2 = 40^2 + 30^2 - 2 \times 40 \times 30 \times \cos(\angle CGD).$

Simplifying:

$196 = 1600 + 900 - 2400 \cos(\angle CGD).$

$2400 \cos(\angle CGD) = 2500 - 196 = 2304.$

$\cos(\angle CGD) = \frac{24}{25}.$

Therefore, we can find $\sin(\angle CGD)$ using the identity $\sin^2 \theta + \cos^2 \theta = 1$ :

$\sin(\angle CGD) = \sqrt{1 - \left(\frac{24}{25}\right)^2} = \frac{7}{25}.$

Now, the area of triangle $CDG$ is

$\frac{1}{2} \times 40 \times 30 \times \frac{7}{25} = 168.$

Noting that the height of triangle $CDG$ is the same as the height of triangle $BGE$ , the ratio of the areas of the two triangles will be the same as the ratio of their corresponding lengths. Therefore, the answer is

$\frac{168 \times 39}{14} = \boxed{468}.$

(Feel free to add or correct any LaTeX and formatting)

~ Mitsuihisashi14, edited by aoum

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2025 AIME II Problems/Problem 1

Contents

Problem

Solution 1

Solution 2 (Law of Cosines)

See also