Competition problem — Geometry — Maths Olympiad Prep

Library / /106 of 860

Geometry Difficulty 4.8 AIME Find the answer

In the figure, if $A E=3, C E=1, B D=C D=2$ , and $A B=5$ , find $A G$ .

Your answer

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

Solution

By Stewart's Theorem, $A D^{2} \cdot B C+C D \cdot B D \cdot B C=A B^{2} \cdot C D+A C^{2} \cdot B D$ , so $A D^{2}=\left(5^{2} \cdot 2+4^{2} \cdot 2-2 \cdot 2 \cdot 4\right) / 4=(50+32-16) / 4=33 / 2$ . By Menelaus's Theorem applied to line $B G E$ and triangle $A C D, D G / G A \cdot A E / E C \cdot C B / B D=1$ , so $D G / G A=1 / 6 \Rightarrow A D / A G=7 / 6$ . Thus $A G=6 \cdot A D / 7=3 \sqrt{66} / 7$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.