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.

Another footnote

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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