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2010 USAMO Problems/Problem 1

Revision as of 04:04, 7 May 2010 by Aopsvd (talk | contribs) (Created page with '==Problem== Let <math>AXYZB</math> be a convex pentagon inscribed in a semicircle of diameter <math>AB</math>. Denote by <math>P, Q, R, S</math> the feet of the perpendiculars fr…')
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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

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; unitsize(1inch); void langle(picture p=currentpicture, pair A, pair B, pair C, string l="", real r=20) { path a = anglemark(A, B, C, r); draw(p, a); pair lp = bisectorpoint(A, B, C); label(p, "$\scriptstyle{" + l + "}$", B + 1.2 * r * markscalefactor * (lp - B)); } picture p; real r = 1.75; pair A = r * plain.W; dot(p, A); label(p, "$A$", A, plain.W); pair B = r * plain.E; dot(p, B); label(p, "$B$", B, plain.E); pair O = (0,0); dot(p, O); label(p, "$O$", O, plain.S); real alpha = 22.5; // angle BAZ real beta = 15; // angble ABX real delta = 30; // angle ZAY real gamma = 90 - alpha - beta - delta; // angle XBY // Points X, Y, Z pair X = r * dir(180-2*beta); dot(p, X); label(p, "$X$", X, plain.WNW); pair Y = r * dir(2*(alpha + delta)); dot(p, Y); label(p, "$Y$", Y, plain.NNW); pair Z = r * dir(2*alpha); dot(p, Z); label(p, "$Z$", Z, plain.NE); langle(p, B, A, Z, "\alpha"); langle(p, X, B, A, "\beta"); langle(p, X, B, A, "\beta"); langle(p, Y, B, X, "\gamma"); langle(p, Y, A, X, "\gamma"); langle(p, Z, A, Y, "\delta"); langle(p, Z, B, Y, "\delta"); // Semi-circle with diameter A--B path C=B..r*plain.N..A--cycle; draw(p, C); pair Q = foot(Y, B, X); dot(p, Q); label(p, "$Q$", Q, plain.SW); draw(p, Y--Q); draw(p, rightanglemark(Y, Q, X, 3)); pair P = foot(Y, A, X); dot(p, P); label(p, "$P$", P, plain.NW); draw(p, A--P); draw(p, Y--P); draw(p, rightanglemark(Y, P, A, 3)); pair S = foot(Y, A, Z); dot(p, S); label(p, "$S$", S, plain.SE); draw(p, Y--S); draw(p, rightanglemark(Z, S, Y, 3)); pair R = foot(Y, B, Z); dot(p, R); label(p, "$R$", R, plain.NE); draw(p, B--R); draw(p, Y--R); draw(p, rightanglemark(B, R, Y, 3)); // Triangles AB{X,Y,Z} draw(p, A--Y); draw(p, A--Z); draw(p, B--X); draw(p, B--Y); draw(p, rightanglemark(B, Y, A, 3)); draw(p, rightanglemark(B, Z, A, 3)); draw(p, rightanglemark(B, X, A, 3)); // Y projection on AB pair T = foot(Y, A, B); dot(p, T); label(p, "$T$", T, plain.S); pen cp = currentpen; currentpen = linewidth(0.3); draw(p, P--T); draw(p, R--T); langle(p, R, T, P, "\chi", 10); currentpen = cp; add(p); [/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, $RY : YS = BY : YA$, and so $\angle YSR = \angle YAB = \alpha + \delta$.

Now $SY$ is perpendicular to $AZ$ so the direction $SY$ is $\alpha$ anti-clockwise from the vertical, and since $\angle YSR = \alpha + \delta$ we see that $SR$ is $\delta$ clockwise from the vertical.

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 $\gamma$ anti-clockwise 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.

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 $YS = AY \sin(\delta)$ and is inclined $\alpha$ anti-clockwise 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 $Y$.

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