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HMMT February

Geometry Difficulty 4.8 AIME Find the answer

In the figure, if AE=3,CE=1,BD=CD=2A E=3, C E=1, B D=C D=2, and AB=5A B=5, find AGA G.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

By Stewart's Theorem, AD2⋅BC+CD⋅BD⋅BC=AB2⋅CD+AC2⋅BDA D^{2} \cdot B C+C D \cdot B D \cdot B C=A B^{2} \cdot C D+A C^{2} \cdot B D, so AD2=(52⋅2+42⋅2−2⋅2⋅4)/4=(50+32−16)/4=33/2A D^{2}=\left(5^{2} \cdot 2+4^{2} \cdot 2-2 \cdot 2 \cdot 4\right) / 4=(50+32-16) / 4=33 / 2. By Menelaus's Theorem applied to line BGEB G E and triangle ACD,DG/GA⋅AE/EC⋅CB/BD=1A C D, D G / G A \cdot A E / E C \cdot C B / B D=1, so DG/GA=1/6⇒AD/AG=7/6D G / G A=1 / 6 \Rightarrow A D / A G=7 / 6. Thus AG=6⋅AD/7=366/7A G=6 \cdot A D / 7=3 \sqrt{66} / 7.

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