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Difference between revisions of "2010 USAMO Problems/Problem 1"

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Note that because <math>\angle{QPY}</math> and <math>\angle{YAB}</math> are both complementary to <math>\beta + \gamma</math>, they must be equal. Now, let <math>PQ</math> intersect diameter <math>AB</math> at <math>T'</math>. Then <math>PYT'A</math> is cyclic and so <math>\angle{YT'A} = 180^\circ - \angle{APY} = 90^\circ</math>. Hence <math>T'YSB</math> is cyclic as well, and so we deduce that <math>\angle{YST'} = \angle{YBT'} = 90^\circ - \alpha - \delta = \angle{YSR}.</math> Hence <math>S, R, T'</math> are collinear and so <math>T = T'</math>. This proves the Footnote.
 
Note that because <math>\angle{QPY}</math> and <math>\angle{YAB}</math> are both complementary to <math>\beta + \gamma</math>, they must be equal. Now, let <math>PQ</math> intersect diameter <math>AB</math> at <math>T'</math>. Then <math>PYT'A</math> is cyclic and so <math>\angle{YT'A} = 180^\circ - \angle{APY} = 90^\circ</math>. Hence <math>T'YSB</math> is cyclic as well, and so we deduce that <math>\angle{YST'} = \angle{YBT'} = 90^\circ - \alpha - \delta = \angle{YSR}.</math> Hence <math>S, R, T'</math> are collinear and so <math>T = T'</math>. This proves the Footnote.
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 +
===Footnote to the Footnote to the Footnote===
 +
The Footnote's claim can be proved even more easily as follows.
 +
 +
Drop an altitude from <math>Y</math> to <math>AB</math> at point <math>T</math>.  Notice that <math>P, Q, T</math> are collinear because they form the Simson line of <math>\triangle AXB</math> from <math>Y</math>.  Also notice that <math>P, Q, T</math> are collinear because they form the Simson line of <math>\triangle AZB</math> from <math>Y</math>.  Since <math>T</math> is at the diameter <math>AB</math>, lines <math>PQ</math> and <math>SR</math> must intersect at the diameter.
  
 
== See Also ==
 
== See Also ==

Revision as of 12:16, 8 March 2018

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; // 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 $YS = AY \sin(\delta)$ 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.

See Also

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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