Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.8 AIME, harder Prove it United States

Problem:

In the following figure—not drawn to scale!— $E$ is the midpoint of $BC$ , triangle $FEC$ has area $7$ , and quadrilateral $DBEG$ has area $27$ . Triangles $ADG$ and $GEF$ have the same area, $x$ . Find $x$ .

Solution

Solution:

The answer is $x=8$ .

Use the notation $[\cdot]$ to denote the area of a polygon. Draw $GB$ ; notice that triangles $GBE$ and $GEC$ have equal bases and altitudes, so $[GBE]=[GEC]=x+7$ . Since $[ABE]=27+x$ , we have $[GDB]=20-x$ .

Likewise, if we draw $AC$ , we see that $[ABE]=[AEC]=27+x$ , so $[AGC]=20$ , which implies that $[CAD]=20+x$ .

Now triangles $GAD$ and $GDB$ have the same altitude (from $G$ to $AB$ ), so their bases are proportional to their respective areas. In other words,
$\frac{AD}{DB}=\frac{[GAD]}{[GDB]}=\frac{x}{20-x}$
But $AD$ and $DB$ are also the bases of triangles $CAD$ and $CDB$ , which have the same altitude (from $C$ to $AB$ ). Hence
$\frac{AD}{DB}=\frac{[CAD]}{[CDB]}=\frac{20+x}{34+x}$
Equating these two fractions leads to the quadratic equation $34x + x^2 = 400 - x^2$ ; the only positive solution is $x=8$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.