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

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<math>O</math> is the midpoint of segment <math>AB</math>.
 
<math>O</math> is the midpoint of segment <math>AB</math>.
  
==Solution==
+
==Solution 1==
 
Let <math>\alpha = \angle BAZ</math>, <math>\beta = \angle ABX</math>.
 
Let <math>\alpha = \angle BAZ</math>, <math>\beta = \angle ABX</math>.
 
Since <math>XY</math> is a chord of the circle with diameter <math>AB</math>,
 
Since <math>XY</math> is a chord of the circle with diameter <math>AB</math>,
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Similarly, <math>RY : YS = AY : YB</math>, and so <math>\angle YRS = \angle YAB = \alpha + \delta</math>.
 
Similarly, <math>RY : YS = AY : YB</math>, and so <math>\angle YRS = \angle YAB = \alpha + \delta</math>.
  
Now <math>RY</math> is perpendicular to <math>AZ</math> so the direction <math>RY</math> is <math>\alpha</math> counterclockwise from the vertical, and since <math>\angle YRS = \alpha + \delta</math> we see that <math>SR</math> is <math>\delta</math> clockwise from the vertical.
+
Now <math>RY</math> is perpendicular to <math>AZ</math> so the direction <math>RY</math> is <math>\alpha</math> counterclockwise from the vertical, and since <math>\angle YRS = \alpha + \delta</math> we see that <math>SR</math> is <math>\delta</math> clockwise from the vertical. (Draw an actual vertical line segment if necessary.)
  
 
Similarly, <math>QY</math> is perpendicular to <math>BX</math> so the direction <math>QY</math> is <math>\beta</math> clockwise from the vertical, and since <math>\angle YQP</math> is <math>\gamma + \beta</math> we see that <math>QY</math> is <math>\gamma</math> counterclockwise from the vertical.
 
Similarly, <math>QY</math> is perpendicular to <math>BX</math> so the direction <math>QY</math> is <math>\beta</math> clockwise from the vertical, and since <math>\angle YQP</math> is <math>\gamma + \beta</math> we see that <math>QY</math> is <math>\gamma</math> counterclockwise from the vertical.
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below the point <math>Y</math>.
 
below the point <math>Y</math>.
  
Since <math>YS = AY \sin(\delta)</math> and is inclined <math>\alpha</math> counterclockwise
+
Since <math>YR = AY \sin(\delta)</math> and is inclined <math>\alpha</math> counterclockwise
 
from the vertical, the point <math>S</math> is <math>AY \sin(\delta) \sin(\alpha)</math>
 
from the vertical, the point <math>S</math> is <math>AY \sin(\delta) \sin(\alpha)</math>
 
horizontally to the right of <math>Y</math>.
 
horizontally to the right of <math>Y</math>.
Line 121: Line 121:
  
 
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.
 +
 +
===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.
 +
 +
===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 <math>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!
 +
 +
==Solution 2==
 +
<asy>
 +
currentpicture=new picture;
 +
size(12cm);
 +
pair O, A, B, X, Y, Z, P, Q, R, SS, T;
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O=(0, 0);
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A=(-1, 0);
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B=(1, 0);
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X=(Cos(144), Sin(144));
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Y=(Cos(105), Sin(105));
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Z=(Cos(27), Sin(27));
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P=foot(Y, A, X);
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Q=foot(Y, B, X);
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R=foot(Y, A, Z);
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SS=foot(Y, B, Z);
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T=foot(Y, A, B);
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dot(O); dot(A); dot(B); dot(X); dot(Y); dot(Z); dot(P); dot(Q); dot(R); dot(SS); dot(T);
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draw(arc(O, 1, 0, 180));
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draw(circumcircle(T, A, Y), dotted);
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draw(circumcircle(T, B, Y), dotted);
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draw(A -- B);
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draw(Z -- O -- X -- A -- Z -- B -- X);
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draw(A -- Y -- B);
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draw(P -- T -- SS);
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draw(P -- Y -- Q); draw(R -- Y -- SS);
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draw(X -- P); draw(Z -- SS);
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draw(Y -- T);
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draw(rightanglemark(Y, T, B, 1.25));
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draw(rightanglemark(Y, P, A, 1.25));
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draw(rightanglemark(Y, Q, X, 1.25));
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draw(rightanglemark(Y, R, Z, 1.25));
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draw(rightanglemark(Y, SS, B, 1.25));
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draw(rightanglemark(A, X, B, 1.25));
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draw(rightanglemark(A, Y, B, 1.25));
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draw(rightanglemark(A, Z, B, 1.25));
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label("$O$", O, S);
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label("$A$", A, SW);
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label("$B$", B, SE);
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label("$X$", X, (X-B)/length(X-B));
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label("$Y$", Y, Y);
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label("$Z$", Z, (Z-A)/length(Z-A));
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label("$P$", P, (P-T)/length(P-T));
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label("$Q$", Q, SW);
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label("$R$", R, SE);
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label("$S$", SS, (SS-T)/length(SS-T));
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label("$T$", T, S);
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</asy>
 +
Let <math>T</math> be the projection of <math>Y</math> onto <math>\overline{AB}</math>. Notice that <math>T</math> lies on the Simson Line <math>\overline{PQ}</math> from <math>Y</math> to <math>\triangle AXB</math>, and the Simson Line <math>\overline{RS}</math> from <math>Y</math> to <math>\triangle AZB</math>. Hence, <math>T=\overline{PQ}\cap\overline{RS}</math>, so it suffices to show that <math>\angle PTS=\tfrac{1}{2}\angle XOZ</math>.
 +
 +
Since <math>TAPY</math> and <math>TBSY</math> are cyclic quadrilaterals, <cmath>\angle PTS=\angle PTY+\angle YTS=\angle PAY+\angle YBS=\angle XAY+\angle YBZ=\frac{1}{2}\angle XOY+\frac{1}{2}\angle YOZ=\frac{1}{2}\angle XOZ,</cmath>
 +
as required. <math>\square</math>
 +
 +
Solution by TheUltimate123.
  
 
== See Also ==
 
== See Also ==

Revision as of 19:09, 27 December 2020

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 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 projection of $Y$ onto $\overline{AB}$. Notice that $T$ lies on the Simson Line $\overline{PQ}$ from $Y$ to $\triangle AXB$, and the Simson Line $\overline{RS}$ from $Y$ to $\triangle AZB$. Hence, $T=\overline{PQ}\cap\overline{RS}$, so it suffices to show that $\angle PTS=\tfrac{1}{2}\angle XOZ$.

Since $TAPY$ and $TBSY$ are cyclic quadrilaterals, \[\angle PTS=\angle PTY+\angle YTS=\angle PAY+\angle YBS=\angle XAY+\angle YBZ=\frac{1}{2}\angle XOY+\frac{1}{2}\angle YOZ=\frac{1}{2}\angle XOZ,\] as required. $\square$

Solution by TheUltimate123.

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

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