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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>
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Claim: <math>H</math> is the reflection of <math>O</math> over the angle bisector of <math>\angle BAC</math> (henceforth 'the' reflection)
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Proof: Let <math>H'</math> be the reflection of <math>O</math>, and let <math>B'</math> be the reflection of <math>B</math>.
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Then reflection takes <math>\angle ABH'</math> to <math>\angle AB'O</math>.
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<math>\Delta ABB'</math> is equilateral, and <math>O</math> lies on the perpendicular bisector of <math>\overline{AB}</math>
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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.
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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.
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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>.
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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>
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Similarly, <math>AC=2s-t</math>
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The ratio <math>\frac{[APQ]}{[ABC]}</math> is <math>\frac{AP AQ}{AB AC} = \frac{s^2}{(s+t)(2s-t)}</math>
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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>
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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 $\triangle{ABC}$ be a non-equilateral, acute triangle with $\angle A=60^{\circ}$, and let $O$ and $H$ denote the circumcenter and orthocenter of $\triangle{ABC}$, respectively.

(a) Prove that line $OH$ intersects both segments $AB$ and $AC$.

(b) Line $OH$ intersects segments $AB$ and $AC$ at $P$ and $Q$, respectively. Denote by $s$ and $t$ the respective areas of triangle $APQ$ and quadrilateral $BPQC$. Determine the range of possible values for $s/t$.

Solution

[asy] import olympiad; unitsize(1inch); pair A,B,C,O,H,P,Q,i1,i2,i3,i4; //define dots A=3*dir(50); B=(0,0); C=right*2.76481496; O=circumcenter(A,B,C); H=orthocenter(A,B,C); i1=2*O-H; i2=2*i1-O; i3=2*H-O; i4=2*i3-H; //These points are for extending PQ. DO NOT DELETE! P=intersectionpoint(i2--i4,A--B); Q=intersectionpoint(i2--i4,A--C); //draw dot(P); dot(Q); draw(P--Q); dot(A); dot(B); dot(C); dot(O); dot(H); draw(A--B--C--cycle); //label label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$P$",P,NW); label("$Q$",Q,NE); label("$O$",O,N); label("$H$",H,N); //change O and H label positions if interfering with other lines to be added //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]

Claim: $H$ is the reflection of $O$ over the angle bisector of $\angle BAC$ (henceforth 'the' reflection)

Proof: Let $H'$ be the reflection of $O$, and let $B'$ be the reflection of $B$.

Then reflection takes $\angle ABH'$ to $\angle AB'O$.

$\Delta ABB'$ is equilateral, and $O$ lies on the perpendicular bisector of $\overline{AB}$

It's well known that $O$ lies strictly inside $ABC$, meaning that $\angle ABH' = \angle AB'O = 30^{\circ},$ 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 $H'$ lies on two altitudes, $H$ is the orthocenter, as desired.

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

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

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

Similarly, $AC=2s-t$

The ratio $\frac{[APQ]}{[ABC]}$ is $\frac{AP AQ}{AB AC} = \frac{s^2}{(s+t)(2s-t)}$ The denominator equals $(1.5s)^2-(.5s-t)^2$ where $.5s-t$ can equal any value in $(-.5s, .5s)$ except $0$. Therefore, the denominator can equal any value in $(2s^2, 9s^2/4)$, and the ratio is any value in $\boxed{\left(\frac{4}{9},\frac{1}{2}\right)}.$

Note: It's easy to show that for any point $H$ on $\verline{PQ}$ (Error compiling LaTeX. Unknown error_msg), Points B and C can be validly defined to make an acute, non-equilateral triangle

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