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

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So <math>\overline{OH}</math> is perpendicular to the angle bisector of <math>\angle OAH</math>, which is the same line as the angle bisector of <math>\angle BAC</math>, meaning that <math>\Delta APQ</math> is equilateral. | So <math>\overline{OH}</math> is perpendicular to the angle bisector of <math>\angle OAH</math>, which is the same line as the angle bisector of <math>\angle BAC</math>, meaning that <math>\Delta APQ</math> is equilateral. | ||

− | Let its side length be <math>s</math>, and let <math>PH=t</math>, where <math>0 < t < s, t \neq s/2</math> because O lies strictly within <math>\angle BAC</math>, as must<math> H</math>, the reflection of <math>O</math>. Also, it's easy to show that if <math>O=H</math> in a general triangle, it's | + | Let its side length be <math>s</math>, and let <math>PH=t</math>, where <math>0 < t < s, t \neq s/2</math> because <math>O</math> lies strictly within <math>\angle BAC</math>, as must <math>H</math>, the reflection of <math>O</math>. Also, it's easy to show that if <math>O=H</math> in a general triangle, it's equilateral, and we <math>\Delta ABC</math> is not equilateral. Hence H is not on the bisector of <math>\angle BAC \implies t \neq s/2</math>. Let <math>\overrightarrow{BH}</math> intersect <math>\overline{AC}</math> at <math>P_B</math>. |

Since <math>\Delta HP_BQ</math> and <math>BP_BA</math> are 30-60-90 triangles, <math>AB=2AP_B=2(s-QP_B)=2(s-HQ/2)=2s-HQ=2s-(s-t)=s+t</math> | Since <math>\Delta HP_BQ</math> and <math>BP_BA</math> are 30-60-90 triangles, <math>AB=2AP_B=2(s-QP_B)=2(s-HQ/2)=2s-HQ=2s-(s-t)=s+t</math> |

## Revision as of 01:55, 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 (since it's acute), 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 lies strictly within , as must , the reflection of . Also, it's easy to show that if in a general triangle, it's equilateral, and we is not equilateral. Hence H is not on the bisector of . 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.