2010 USAMO Problems/Problem 1

Problem

Let $AXYZB$ be a convex pentagon inscribed in a semicircle of diameter $AB$ . Denote by $P, Q, R, S$ the feet of the perpendiculars from $Y$ onto lines $AX, BX, AZ, BZ$ , respectively. Prove that the acute angle formed by lines $PQ$ and $RS$ is half the size of $\angle XOZ$ , where $O$ is the midpoint of segment $AB$ .

Solution 0 (play like a God)

Angles are directed mod 180 {don't worry}(take so that the solution can be applied to any configuration) Let the intersection of pq and rs is M.

$m\angle ARM=m\angle ZRS =m \angle ZYS =m\angle YZR = m\angle YZA = m \angle YXP = m \angle XYQ =m \angle XPQ = m \angle APM$ This implies APRM is cyclic So: $m \angle PMR = m \angle PAR =m \angle XAZ = \frac{m\angle XOZ}{2}$ and we are done. For any concern mail on harshsahu1098@gmail.com

Solution 1

Let $\alpha = \angle BAZ$ , $\beta = \angle ABX$ . Since $XY$ is a chord of the circle with diameter $AB$ , $\angle XAY = \angle XBY = \gamma$ . From the chord $YZ$ , we conclude $\angle YAZ = \angle YBZ = \delta$ .

$[asy] import olympiad; // Scale unitsize(1inch); real r = 1.75; // Semi-circle: centre O, radius r, diameter A--B. pair O = (0,0); dot(O); label("$O$", O, plain.S); pair A = r * plain.W; dot(A); label("$A$", A, unit(A)); pair B = r * plain.E; dot(B); label("$B$", B, unit(B)); draw(arc(O, r, 0, 180)--cycle); // points X, Y, Z real alpha = 22.5; real beta = 15; real delta = 30; pair X = r * dir(180 - 2*beta); dot(X); label("$X$", X, unit(X)); pair Y = r * dir(2*(alpha + delta)); dot(Y); label("$Y$", Y, unit(Y)); pair Z = r * dir(2*alpha); dot(Z); label("$Z$", Z, unit(Z)); // Feet of perpendiculars from Y pair P = foot(Y, A, X); dot(P); label("$P$", P, unit(P-Y)); dot(P); pair Q = foot(Y, B, X); dot(P); label("$Q$", Q, unit(A-Q)); dot(Q); pair R = foot(Y, B, Z); dot(R); label("$S$", R, unit(R-Y)); dot(R); pair S = foot(Y, A, Z); dot(S); label("$R$", S, unit(B-S)); dot(S); pair T = foot(Y, A, B); dot(T); label("$T$", T, unit(T-Y)); dot(T); // Segments draw(B--X); draw(B--Y); draw(B--R); draw(A--Z); draw(A--Y); draw(A--P); draw(Y--P); draw(Y--Q); draw(Y--R); draw(Y--S); draw(R--T); draw(P--T); // Right angles draw(rightanglemark(A, X, B, 3)); draw(rightanglemark(A, Y, B, 3)); draw(rightanglemark(A, Z, B, 3)); draw(rightanglemark(A, P, Y, 3)); draw(rightanglemark(Y, R, B, 3)); draw(rightanglemark(Y, S, A, 3)); draw(rightanglemark(B, Q, Y, 3)); // Acute angles import markers; void langle(pair A, pair B, pair C, string l="", real r=40, int n=1, int nm = 0) { string sl = "$\scriptstyle{" + l + "}$"; marker m = (nm > 0) ? marker(markinterval(stickframe(n=nm, 2mm), true)) : nomarker; markangle(Label(sl), radius=r, n=n, A, B, C, m); } langle(B, A, Z, "\alpha" ); langle(X, B, A, "\beta", n=2); langle(Y, A, X, "\gamma", nm=1); langle(Y, B, X, "\gamma", nm=1); langle(Z, A, Y, "\delta", nm=2); langle(Z, B, Y, "\delta", nm=2); langle(R, S, Y, "\alpha+\delta", r=23); langle(Y, Q, P, "\beta+\gamma", r=23); langle(R, T, P, "\chi", r=15); [/asy]$

Triangles $BQY$ and $APY$ are both right-triangles, and share the angle $\gamma$ , therefore they are similar, and so the ratio $PY : YQ = AY : YB$ . Now by Thales' theorem the angles $\angle AXB = \angle AYB = \angle AZB$ are all right-angles. Also, $\angle PYQ$ , being the fourth angle in a quadrilateral with 3 right-angles is again a right-angle. Therefore $\triangle PYQ \sim \triangle AYB$ and $\angle YQP = \angle YBA = \gamma + \beta$ . Similarly, $RY : YS = AY : YB$ , and so $\angle YRS = \angle YAB = \alpha + \delta$ .

Now $RY$ is perpendicular to $AZ$ so the direction $RY$ is $\alpha$ counterclockwise from the vertical, and since $\angle YRS = \alpha + \delta$ we see that $SR$ is $\delta$ clockwise from the vertical. (Draw an actual vertical line segment if necessary.)

Similarly, $QY$ is perpendicular to $BX$ so the direction $QY$ is $\beta$ clockwise from the vertical, and since $\angle YQP$ is $\gamma + \beta$ we see that $QY$ is $\gamma$ counterclockwise from the vertical.

Therefore the lines $PQ$ and $RS$ intersect at an angle $\chi = \gamma + \delta$ . Now by the central angle theorem $2\gamma = \angle XOY$ and $2\delta = \angle YOZ$ , and so $2(\gamma + \delta) = \angle XOZ$ , and we are done.

Note that $RTQY$ is a quadrilateral whose angles sum to 360°; can you find a faster approach using this fact?

Footnote

We can prove a bit more. Namely, the extensions of the segments $RS$ and $PQ$ meet at a point on the diameter $AB$ that is vertically below the point $Y$ .

Since $YR = AY \sin(\delta)$ and is inclined $\alpha$ counterclockwise from the vertical, the point $S$ is $AY \sin(\delta) \sin(\alpha)$ horizontally to the right of $Y$ .

Now $AS = AY \cos(\delta)$ , so $S$ is $AS \sin(\alpha) = AY \cos(\delta)\sin(\alpha)$ vertically above the diameter $AB$ . Also, the segment $SR$ is inclined $\delta$ clockwise from the vertical, so if we extend it down from $S$ towards the diameter $AB$ it will meet the diameter at a point which is $AY \cos(\delta)\sin(\alpha)\tan(\delta) = AY \sin(\delta)\sin(\alpha)$ horizontally to the left of $S$ . This places the intersection point of $RS$ and $AB$ vertically below $Y$ .

Similarly, and by symmetry the intersection point of $PQ$ and $AB$ is directly below $Y$ on $AB$ , so the lines through $PQ$ and $RS$ meet at a point $T$ on the diameter that is vertically below $Y$ .

Footnote to the Footnote

The Footnote's claim is more easily proved as follows.

Note that because $\angle{QPY}$ and $\angle{YAB}$ are both complementary to $\beta + \gamma$ , they must be equal. Now, let $PQ$ intersect diameter $AB$ at $T'$ . Then $PYT'A$ is cyclic and so $\angle{YT'A} = 180^\circ - \angle{APY} = 90^\circ$ . Hence $T'YSB$ is cyclic as well, and so we deduce that $\angle{YST'} = \angle{YBT'} = 90^\circ - \alpha - \delta = \angle{YSR}.$ Hence $S, R, T'$ are collinear and so $T = T'$ . This proves the Footnote.

Footnote to the Footnote to the Footnote

The Footnote's claim can be proved even more easily as follows.

Drop an altitude from $Y$ to $AB$ at point $T$ . Notice that $P, Q, T$ are collinear because they form the Simson line of $\triangle AXB$ from $Y$ . Also notice that $P, Q, T$ are collinear because they form the Simson line of $\triangle AZB$ from $Y$ . Since $T$ is at the diameter $AB$ , lines $PQ$ and $SR$ must intersect at the diameter.

Footnote to the Fourth Power

There is another, simpler solution using Simson lines. Can you find it?

Operation Diagram

Of course, as with any geometry problem, DRAW A HUGE DIAGRAM spanning at least one page. And label all your right angles, noting rectangles $PXQY$ and $YSZR$ . It looks like there are a couple of key angles we need to diagram. Let's take $\angle{ZAB} = \alpha, \angle{XBA} = \beta, \angle{YAZ} = \angle{YBZ} = \delta$ . From there $\angle{XOZ}=180^\circ - \angle{XOA}-\angle{ZOB}=180-2(\beta + \alpha)$ .

Move on to the part about the intersection of $PQ$ and $RS$ . Call the intersection $J$ . Note that by Simson Lines from point $Y$ to $\triangle{ABX}$ and $\triangle{AZB}$ , $YJ$ is perpendicular to $AB$ and $J$ lies on $AB$ . Immediately note that we are trying to show that $\angle{PJS} = 90 - \beta - \alpha$ .

It suffices to show that referencing quadrilateral $QR~J$ , where $~$ represents the intersection of $XB, AZ$ , we have reflex $\angle{Q~R} + \angle{BQJ} + \angle{ARJ} = 270 + \alpha + \beta$ . Note that the reflex angle is $180^\circ + \angle{A~X} = 180^\circ + (90^\circ - \angle{XA*}) = 270^\circ - ((90 - \beta) - \alpha) = 180 ^\circ + \alpha + \beta$ , therefore it suffices to show that $\angle{BQJ} + \angle{ARJ} = 90^\circ$ . To make this proof more accessible, note that via (cyclic) rectangles $PXQY$ and $YSZR$ , it suffices to prove $\angle{YPJ} + \angle{YSJ} = 90^\circ$ .

Note $\angle{YPJ} = \angle{YPQ} = \angle{YXQ} = \angle{YXB} = \angle{YAB} = \alpha + \delta$ . Note $\angle{YSJ} = \angle{YSR} = \angle{YZR} = \angle{YZA} = \angle{YBA} = \angle{YBX} + \angle{XBA} = ((90^\circ - \alpha) - \delta - \beta) + \beta = 90^\circ - \alpha - \delta$ , which completes the proof.

Footnote to Operation Diagram

For reference/feasibility records: took expiLnCalc ~56 minutes (consecutively). During the problem expiLnCalc realized that the inclusion of $\delta$ was necessary when trying to show that $\angle{YSJ}+\angle{YPJ}=90^\circ$ . Don't be afraid to attempt several different strategies, and always be humble!

Solution 2

$[asy] currentpicture=new picture; size(12cm); pair O, A, B, X, Y, Z, P, Q, R, SS, T; O=(0, 0); A=(-1, 0); B=(1, 0); X=(Cos(144), Sin(144)); Y=(Cos(105), Sin(105)); Z=(Cos(27), Sin(27)); P=foot(Y, A, X); Q=foot(Y, B, X); R=foot(Y, A, Z); SS=foot(Y, B, Z); T=foot(Y, A, B); dot(O); dot(A); dot(B); dot(X); dot(Y); dot(Z); dot(P); dot(Q); dot(R); dot(SS); dot(T); draw(arc(O, 1, 0, 180)); draw(circumcircle(T, A, Y), dotted); draw(circumcircle(T, B, Y), dotted); draw(A -- B); draw(Z -- O -- X -- A -- Z -- B -- X); draw(A -- Y -- B); draw(P -- T -- SS); draw(P -- Y -- Q); draw(R -- Y -- SS); draw(X -- P); draw(Z -- SS); draw(Y -- T); draw(rightanglemark(Y, T, B, 1.25)); draw(rightanglemark(Y, P, A, 1.25)); draw(rightanglemark(Y, Q, X, 1.25)); draw(rightanglemark(Y, R, Z, 1.25)); draw(rightanglemark(Y, SS, B, 1.25)); draw(rightanglemark(A, X, B, 1.25)); draw(rightanglemark(A, Y, B, 1.25)); draw(rightanglemark(A, Z, B, 1.25)); label("$O$", O, S); label("$A$", A, SW); label("$B$", B, SE); label("$X$", X, (X-B)/length(X-B)); label("$Y$", Y, Y); label("$Z$", Z, (Z-A)/length(Z-A)); label("$P$", P, (P-T)/length(P-T)); label("$Q$", Q, SW); label("$R$", R, SE); label("$S$", SS, (SS-T)/length(SS-T)); label("$T$", T, S); [/asy]$

Let $T$ be the foot of the perpendicular from $Y$ to $\overline{AB}$ , let $O$ be the center of the semi-circle.

Since we have a semi-circle, if we were to reflect it over $\overline {AB}$ , we would have a full circle, with $\triangle{AXB}$ and $\triangle{AZB}$ inscribed in it. Now, notice that $Y$ is a point on that full circle, so we can say that $T$ lies on the Simson Line $\overline{PQ}$ from $Y$ to $\triangle AXB$ and that it also lies on the Simson line $\overline {RS}$ from $Y$ to $\triangle AZB$ . Thus, $T$ lies on two distinct lines in a plane, which means that $T=\overline{PQ}\cap\overline{RS}$ . Therefore, it suffices to show that $\angle PTS=\tfrac{1}{2}\angle XOZ$ .

Since $m\angle YTA + m\angle YPA = 90^\circ + 90^\circ = 180^\circ$ and $m \angle YTB + m \angle YSB = 90^\circ + 90^\circ = 180^\circ$ , we know that $TAPY$ and $TBSY$ are cyclic quadrilaterals.

We use this fact to get $\angle PTS=\angle PTY+\angle YTS=\angle PAY+\angle YBS=\angle XAY+\angle YBZ. \space \space (1)$ \\ Now note that $\angle XAY$ is the inscribed angle of minor arc $\overset{\huge\frown}{PY}$ , and $\angle XOY$ is the central angle of minor arc $\overset{\huge\frown}{AB}$ , so $\angle XAY = \frac{\overset{\huge\frown}{PY}}{2} = \frac{\angle XOY}{2}$ . Similarly, $\angle YBZ = \frac{\overset{\huge\frown}{YZ}}{2}=\frac{\angle YOZ}{2}$ . Thus we can say $\angle XAY + \angle YBZ = \frac{\angle XOY}{2} + \frac{\angle YOZ}{2}=\frac{\angle XOY + \angle YOZ}{2} = \frac{\angle XOZ}{2}. \space \space (2)$

Combining statements $(1)$ and $(2)$ , we can say that $\angle PTS = \frac{\angle XOZ}{2}$ , as desired. $\square$

~thinker123

2010 USAMO (Problems • Resources)
Preceded by First problem	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

2010 USAJMO (Problems • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6
All USAJMO Problems and Solutions

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