Problem Post 2-2: A Sampler of `Olympiad' Geometry Concepts

Problem Posts

[Level 4]
Henry

This post is meant to be a `sampler' of sorts, to show the most common tag words one will see in olympiad geometry problems.

I've been fortunate enough to find two remarkable problems, and solve them in ways that form a whirlwind tour of the subject. The first problem demonstrates side chasing, isosceles triangles, some cyclic quadrilaterals and spiral similarity. The second problem demonstrates power theorems, cyclic quadrilaterals, collinearity, and triangle geometry.

Problem 1

This one comes from Andreescu and Gelca's Mathematical Olympiad Challenges.

QUESTION
(Andreescu, Gelca) Let $B$ and $C$ be the endpoints and $A$ the midpoint of a semicircle. Let $M$ be a point on the line segment $AC$, and $P$, $Q$ the feet of the perpendiculars from $A$ and $C$ to the line $BM$, respectively. Prove that $BP=PQ+QC$.

Problem 1-6 Romance of a Right Triangle

Problem Posts

[Difficulty 4] [PDF]

QUESTION
(Canadian Mathematical Olympiad 2013, Problem 3) Let $G$ be the centroid of a right-angled triangle $ABC$ with $\angle BCA = 90^{\circ}$. Let $P$ be the point on ray $AG$ such that $\angle CPA = \angle CAB$, and let $Q$ be the point on ray $BG$ such that $\angle CQB = \angle ABC$. Prove that the circumcircles of triangles $AQG$ and $BPG$ meet at a point on side $AB$.

Laconic Solution Sketch.
Using similar triangles, we can show that this common point is the foot of the altitude on $AB$ from $C$.