2014 USAJMO Problems/Problem 2

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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 assistance and are not to be used [/asy]

Lemma: $H$ 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 know $\Delta ABC$ 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.

Solution 2

Let $J$ be the farthest point on the circumcircle of $\triangle{ABC}$ from line $BC$. Lemma: Line $AJ$||Line $OH$ Proof: Set $b=-\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}$ and $c=-\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}$, and $a$ on the unit circle. It is well known that $o=0$ and $h=a+b+c$, so we have $h-o=a+b+c=a-1=a-j$, so $\dfrac{a-j}{h-o}$ is real and thus the 2 lines are parallel.

WLOG let $A$ be in the first quadrant. Clearly by the above lemma $OH$ must intersect line $BC$ closer to $B$ than to $C$. Intersect $AJ$ and $BC$ at $D$ and $OH$ and $BC$ at $E$. We clearly have $0 < \dfrac{\overarc{JC}-\overarc{AB}}{2} = \dfrac{120-\overarc{AC}}{2} = \angle{ADB} = \angle{OEC} < 30 = \angle{OBC}$, $OH$ must intersect $AB$. We also have, letting the intersection of line $OH$ and line $AC$ be $Q$, and letting intersection of $OH$ and $AB$ be $P$, $\angle{OQA} = \angle{JAB} = \dfrac{\overarc{JB}}{2} = 60$. Since $\angle{OCA}=\angle{BCA}-\angle{BCO} < 90-30 =\angle{OQA}$, and $\angle{OQC} = 120>\angle{OAC}$, $OH$ also intersects $AC$. We have $\angle{OPA}=\angle{PAJ}=60$, so $\triangle{APQ}$ is equilateral. Letting $AJ=2x$, and letting the foot of the perpendicular from $O$ to $AJ$ be $L$, we have $OL=\sqrt{1-x^{2}}$, and since $OL$ is an altitude of $\triangle{APQ}$, we have $[APQ]=\dfrac{OL^{2}\sqrt{3}}{3} = \dfrac{\sqrt{3}(1-x^{2})}{3}$. Letting the foot of the perpendicular from $A$ to $OJ$ be $K$, we have $\triangle{JKA}\sim \triangle{JLO}$ by AA with ratio $\dfrac{JA}{JO} = 2x$. Therefore, $JK = 2x(JL) = 2x^{2}$. Letting $D$ be the foot of the altitude from $J$ to $BC$, we have $KD=JD-JK=\dfrac{3}{2}-2x^{2}$, since $re(B)=re(C) \implies JD=re(j)-(-\dfrac{1}{2})=\dfrac{3}{2}$. Thus, since $BC=\sqrt{3}$ we have $[ABC]=\dfrac{(\dfrac{3}{2}-2x^{2})(\sqrt{3})}{2} = \dfrac{(3-4x^{2})(\sqrt{3})}{2}$, so $[PQCB]=[ABC]-[APQ]=\dfrac{(5-8x^{2})(\sqrt{3})}{12}$, so $\dfrac{[APQ]}{[PQCB]} = \dfrac{4-4x^{2}}{5-8x^{2}} = \dfrac{1}{2}+\dfrac{3}{10-16x^{2}}$. We have $x=\sin(\dfrac{JOA}{2})$, with $0<120-2\angle{ACB}=\angle{JOA}<60$, so $x$ can be anything in the interval $(0, \dfrac{1}{2})$. Therefore, the desired range is $(\dfrac{4}{5}, 1)$.

Solution by Shaddoll

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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