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

Stevenmeow (talk | contribs) (→Solution) |
Stevenmeow (talk | contribs) (→Solution) |
||

Line 61: | Line 61: | ||

<math>\Delta ABB'</math> is equilateral, and <math>O</math> lies on the perpendicular bisector of <math>\overline{AB}</math> | <math>\Delta ABB'</math> is equilateral, and <math>O</math> lies on the perpendicular bisector of <math>\overline{AB}</math> | ||

− | It's well known that <math>O</math> lies strictly inside <math>ABC</math>, meaning that <math>\angle ABH' = \angle AB'O = 30^{\circ},</math> from which it follows that <math>\overline{BH'} \perp \verline{AC}</math> . Similarly, <math>\overline{CH'} \perp \verline{AB}</math>. Since <math>H'</math> lies on two altitudes, <math>H</math> is the orthocenter, as desired. | + | It's well known that <math>O</math> lies strictly inside <math>\Delta ABC</math> (since it's acute), meaning that <math>\angle ABH' = \angle AB'O = 30^{\circ},</math> from which it follows that <math>\overline{BH'} \perp \verline{AC}</math> . Similarly, <math>\overline{CH'} \perp \verline{AB}</math>. Since <math>H'</math> lies on two altitudes, <math>H</math> is the orthocenter, as desired. |

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. |

## 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 (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 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.