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

(Added new solution.)
Line 129: Line 129:
 
===Another footnote===
 
===Another footnote===
 
There is another, more simpler solution using Simson lines. Can you find it?
 
There is another, more 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 <math>rectangles PXQY</math> and <math>YSZR</math>. It looks like there are a couple of key angles we need to diagram. Let's take <math>\angle{ZAB} = \alpha, \angle{XBA} = \beta, \angle{YAZ} = \angle{YBZ} = \delta</math>. From there <math>\angle{XOZ}=180^\circ - \angle{XOA}-\angle{ZOB}=180-2(\beta + \alpha)</math>.
 +
 +
Move on to the part about the intersection of <math>PQ</math> and <math>RS</math>. Call the intersection <math>J</math>. Note that by Simson Lines from point <math>Y</math> to <math>\triangle{ABX}</math> and <math>\triangle{AZB}</math>, <math>YJ</math> is perpendicular to <math>AB</math> and <math>J</math> lies on <math>AB</math>. Immediately note that we are trying to show that <math>\angle{PJS} = 90 - \beta - \alpha</math>.
 +
 +
It suffices to show that referencing quadrilateral <math>QR*J</math>, where <math>*</math> represents the intersection of <math>XB, AZ</math>, we have reflex <math>\angle{Q*R} + \angle{BQJ} + \angle{ARJ} = 270 + \alpha + \beta</math>. Note that the reflex angle is <math>180^\circ + \angle{A*X} = 180^\circ + (90^\circ - \angle{XA*}) = 270^\circ - ((90 - \beta) - \alpha) = 180 ^\circ + \alpha + \beta</math>, therefore it suffices to show that <math>\angle{BQJ} + \angle{ARJ} = 90^\circ</math>. To make this proof more accessible, note that via (cyclic) rectangles <math>PXQY</math> and <math>YSZR</math>, it suffices to prove <math>\angle{YPJ} + \angle{YSJ} = 90^\circ</math>.
 +
 +
Note <math>\angle{YPJ} = \angle{YPQ} = \angle{YXQ} = \angle{YXB} = \angle{YAB} = \alpha + \delta</math>.
 +
Note <math>\angle{YSJ} = \angle{YSR} = \angle{YZR} = \angle{YZA} = \angle{YBA} = \angle{YBX} + \angle{XBA} = ((90^\circ - \alpha) - \delta - \beta) + \beta = 90^\circ - \alpha - \delta</math>, 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 <math>\delta</math> was necessary when trying to show that <math>\angle{YSJ}+\angle{YPJ}=90^\circ</math>. Don't be afraid to attempt several different strategies, and always be humble!
  
 
== See Also ==
 
== See Also ==

Revision as of 22:43, 5 April 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.

Another footnote

There is another, more 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!

See Also

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png