A triangle $OAB$ with $\angle A=90^{\circ}$ lies inside another triangle with vertex $O$. The altitude of $OAB$ from $A$ until it meets the side of angle $O$ at $M$. The distances from $M$ and $B$ to the second side of angle $O$ are $2$ and $1$ respectively. Find the length of $OA$.
Problem
Source: Sharygin 2019 Finals Day 1 Grade 9 P1
Tags: geometry, Sharygin Geometry Olympiad
03.09.2019 17:46
MathPassionForever wrote: A triangle $OAB$ with $\angle A=90^{\circ}$ lies inside another triangle with vertex $O$. The altitude of $OAB$ from $A$ until it meets the side of angle $O$ at $M$. The distances from $M$ and $B$ to the second side of angle $O$ are $2$ and $1$ respectively. Find the length of $OA$. The wording should be corrected as follows: A triangle $OAB$ with $\angle A=90^{\circ}$ lies inside a right angle with vertex $O$. The altitude of $OAB$ from $A$ is extended beyond $A$ until it meets the side of angle $O$ at $M$. The distances from $M$ and $B$ to the second side of angle $O$ are $2$ and $1$ respectively. Find the length of $OA$. (here sides of an angle refer to its rays)
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03.09.2019 18:01
Let $\angle COD$ be that right angle with vertex $O$. Let $P, Q$ be the distances from $B$ to $OC, OD$, respectively. By the conditions, $MO=2, BP=1$. Since $BPOQ$ is rectangle, $QO=BP=1$. Let $A_1$ be the foot of the altitude from $A$ in the right triangle $OAB$. By Euclid's Laws, we have $OA^2=OA_1\cdot OB$. But $\angle BA_1M=90^\circ=\angle BQM$ implies that $BA_1QM$ is cyclic, so by power of a point, we get $OA_1\cdot OB=OQ\cdot OM$. Therefore, $OA^2=1\cdot 2=2$, i.e. $OA=\sqrt{2} \,\blacksquare$
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