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

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//further editing: ABCPQOH are the dots to be further used. i1,i2,i3,i4 are for drawing assistence and are not to be used | //further editing: ABCPQOH are the dots to be further used. i1,i2,i3,i4 are for drawing assistence and are not to be used | ||

</asy> | </asy> | ||

+ | |||

+ | Claim: <math>H</math> is the reflection of <math>O</math> over the angle bisector of <math>\angle BAC</math> (henceforth 'the' reflection) | ||

+ | |||

+ | Proof: Let <math>H'</math> be the reflection of <math>O</math>, and let <math>B'</math> be the reflection of <math>B</math>. | ||

+ | |||

+ | Then reflection takes <math>\angle ABH'</math> to <math>\angle AB'O</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. | ||

+ | |||

+ | 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 equlateral, and we know that <math>\Delta ABC</math> is not equilateral. Hence <math>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> | ||

+ | |||

+ | Similarly, <math>AC=2s-t</math> | ||

+ | |||

+ | The ratio <math>\frac{[APQ]}{[ABC]}</math> is <math>\frac{AP AQ}{AB AC} = \frac{s^2}{(s+t)(2s-t)}</math> | ||

+ | The denominator equals <math>(1.5s)^2-(.5s-t)^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>(2s^2, 9s^2/4)</math>, and the ratio is any value in <math>\boxed{\left(\frac{4}{9},\frac{1}{2}\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 |

## Revision as of 01:52, 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.), Points B and C can be validly defined to make an acute, non-equilateral triangle