# Difference between revisions of "2014 USAJMO Problems/Problem 2"

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The denominator equals <math>(1.5s)^2-(.5s-t)^2-s^2</math> where <math>.5s-t</math> can equal any value in <math>(-.5s, .5s)</math> except <math>0</math>. Therefore, the denominator can equal any value in <math>(s^2, 5s^2/4)</math>, and the ratio is any value in <math>\boxed{\left(\frac{4}{5},1\right)}.</math> | The denominator equals <math>(1.5s)^2-(.5s-t)^2-s^2</math> where <math>.5s-t</math> can equal any value in <math>(-.5s, .5s)</math> except <math>0</math>. Therefore, the denominator can equal any value in <math>(s^2, 5s^2/4)</math>, and the ratio is any value in <math>\boxed{\left(\frac{4}{5},1\right)}.</math> | ||

− | Note: It's easy to show that for any point <math>H</math> on <math>\verline{PQ}</math>, Points B and C can be validly defined to make an acute, non-equilateral triangle | + | Note: It's easy to show that for any point <math>H</math> on <math>\verline{PQ}</math> except the midpoint, Points B and C can be validly defined to make an acute, non-equilateral triangle. |

## Revision as of 01:54, 10 May 2014

## 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. ! Undefined control sequence.) . Similarly, $\overline{CH'} \perp \verline{AB}$ (Error compiling LaTeX. ! Undefined control sequence.). 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. ! Undefined control sequence.) except the midpoint, Points B and C can be validly defined to make an acute, non-equilateral triangle.