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>, | ||
Line 143: | Line 143: | ||
For reference/feasibility records: took expiLnCalc ~56 minutes (consecutively). | 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! | 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; | ||
+ | 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("<math>O</math>", O, S); | ||
+ | label("<math>A</math>", A, SW); | ||
+ | label("<math>B</math>", B, SE); | ||
+ | label("<math>X</math>", X, (X-B)/length(X-B)); | ||
+ | label("<math>Y</math>", Y, Y); | ||
+ | label("<math>Z</math>", Z, (Z-A)/length(Z-A)); | ||
+ | label("<math>P</math>", P, (P-T)/length(P-T)); | ||
+ | label("<math>Q</math>", Q, SW); | ||
+ | label("<math>R</math>", R, SE); | ||
+ | label("<math>S</math>", SS, (SS-T)/length(SS-T)); | ||
+ | label("<math>T</math>", T, S); | ||
+ | [/asy] | ||
+ | Let <math>T</math> be the projection of <math>Y</math> onto <math>\seg{AB}</math>. Notice that <math>T</math> lies on the Simson Line <math>\seg{PQ}</math> from <math>Y</math> to <math>\triangle AXB</math>, and the Simson Line <math>\seg{RS}</math> from <math>Y</math> to <math>\triangle AZB</math>. Hence, <math>T=\seg{PQ}\cap\seg{RS}</math>, so it suffices to show that <math>\angle PTS=\frac{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 18:10, 15 August 2018
Contents
Problem
Let be a convex pentagon inscribed in a semicircle of diameter . Denote by the feet of the perpendiculars from onto lines , respectively. Prove that the acute angle formed by lines and is half the size of , where is the midpoint of segment .
Solution 1
Let , . Since is a chord of the circle with diameter , . From the chord , we conclude .
Triangles and are both right-triangles, and share the angle , therefore they are similar, and so the ratio . Now by Thales' theorem the angles are all right-angles. Also, , being the fourth angle in a quadrilateral with 3 right-angles is again a right-angle. Therefore and . Similarly, , and so .
Now is perpendicular to so the direction is counterclockwise from the vertical, and since we see that is clockwise from the vertical. (Draw an actual vertical line segment if necessary.)
Similarly, is perpendicular to so the direction is clockwise from the vertical, and since is we see that is counterclockwise from the vertical.
Therefore the lines and intersect at an angle . Now by the central angle theorem and , and so , and we are done.
Note that 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 and meet at a point on the diameter that is vertically below the point .
Since and is inclined counterclockwise from the vertical, the point is horizontally to the right of .
Now , so is vertically above the diameter . Also, the segment is inclined clockwise from the vertical, so if we extend it down from towards the diameter it will meet the diameter at a point which is horizontally to the left of . This places the intersection point of and vertically below .
Similarly, and by symmetry the intersection point of and is directly below on , so the lines through and meet at a point on the diameter that is vertically below .
Footnote to the Footnote
The Footnote's claim is more easily proved as follows.
Note that because and are both complementary to , they must be equal. Now, let intersect diameter at . Then is cyclic and so . Hence is cyclic as well, and so we deduce that Hence are collinear and so . 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 to at point . Notice that are collinear because they form the Simson line of from . Also notice that are collinear because they form the Simson line of from . Since is at the diameter , lines and 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 and . It looks like there are a couple of key angles we need to diagram. Let's take . From there .
Move on to the part about the intersection of and . Call the intersection . Note that by Simson Lines from point to and , is perpendicular to and lies on . Immediately note that we are trying to show that .
It suffices to show that referencing quadrilateral , where represents the intersection of , we have reflex . Note that the reflex angle is , therefore it suffices to show that . To make this proof more accessible, note that via (cyclic) rectangles and , it suffices to prove .
Note . Note , 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 was necessary when trying to show that . 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, S); label("", A, SW); label("", B, SE); label("", X, (X-B)/length(X-B)); label("", Y, Y); label("", Z, (Z-A)/length(Z-A)); label("", P, (P-T)/length(P-T)); label("", Q, SW); label("", R, SE); label("", SS, (SS-T)/length(SS-T)); label("", T, S); [/asy] Let be the projection of onto $\seg{AB}$ (Error compiling LaTeX. ! Undefined control sequence.). Notice that lies on the Simson Line $\seg{PQ}$ (Error compiling LaTeX. ! Undefined control sequence.) from to , and the Simson Line $\seg{RS}$ (Error compiling LaTeX. ! Undefined control sequence.) from to . Hence, $T=\seg{PQ}\cap\seg{RS}$ (Error compiling LaTeX. ! Undefined control sequence.), so it suffices to show that .
Since and are cyclic quadrilaterals, as required.
Solution by TheUltimate123.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.