2014 USAJMO Problems/Problem 2
Problem
Let be a non-equilateral, acute triangle with , and let and denote the circumcenter and orthocenter of , respectively.
(a) Prove that line intersects both segments and .
(b) Line intersects segments and at and , respectively. Denote by and the respective areas of triangle and quadrilateral . Determine the range of possible values for .
Solution
Claim: is the reflection of over the angle bisector of (henceforth 'the' reflection)
Proof: Let be the reflection of , and let be the reflection of .
Then reflection takes to .
is equilateral, and lies on the perpendicular bisector of
It's well known that lies strictly inside , meaning that from which it follows that $\overline{BH'} \perp \verline{AC}$ (Error compiling LaTeX. Unknown error_msg) . Similarly, $\overline{CH'} \perp \verline{AB}$ (Error compiling LaTeX. Unknown error_msg). Since lies on two altitudes, is the orthocenter, as desired.
So is perpendicular to the angle bisector of , which is the same line as the angle bisector of , meaning that is equilateral.
Let its side length be , and let , where because O lies strictly within , as must, the reflection of . Also, it's easy to show that if in a general triangle, it's equlateral, and we know that is not equilateral. Hence . Let intersect at .
Since and are 30-60-90 triangles,
Similarly,
The ratio is The denominator equals where can equal any value in except . Therefore, the denominator can equal any value in , and the ratio is any value in
Note: It's easy to show that for any point on $\verline{PQ}$ (Error compiling LaTeX. Unknown error_msg), Points B and C can be validly defined to make an acute, non-equilateral triangle