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

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Note: It's easy to show that for any point <math>H</math> on <math>\overline{PQ}</math> except the midpoint, 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>\overline{PQ}</math> except the midpoint, Points B and C can be validly defined to make an acute, non-equilateral triangle.
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==See Also==
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{{USAJMO box|year=2014|num-b=1|num-a=3}}

Revision as of 09:38, 10 March 2015

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 $\Delta ABC$ (since it's acute), meaning that $\angle ABH' = \angle AB'O = 30^{\circ},$ from which it follows that $\overline{BH'} \perp \overline{AC}$ . Similarly, $\overline{CH'} \perp \overline{AB}$. 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 equilateral, and we $\Delta ABC$ is not equilateral. Hence H is not on the bisector of $\angle BAC \implies t \neq s/2$. 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 $\frac{[APQ]}{[ABC]-[APQ]}$ is $\frac{AP \cdot AQ}{AB \cdot AC - AP \cdot AQ} = \frac{s^2}{(s+t)(2s-t)-s^2}$ The denominator equals $(1.5s)^2-(.5s-t)^2-s^2$ where $.5s-t$ can equal any value in $(-.5s, .5s)$ except $0$. Therefore, the denominator can equal any value in $(s^2, 5s^2/4)$, and the ratio is any value in $\boxed{\left(\frac{4}{5},1\right)}.$

Note: It's easy to show that for any point $H$ on $\overline{PQ}$ except the midpoint, Points B and C can be validly defined to make an acute, non-equilateral triangle.

See Also

2014 USAJMO (ProblemsResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6
All USAJMO Problems and Solutions
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