# Difference between revisions of "2013 USAJMO Problems/Problem 5"

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<cmath>\frac{\tan \alpha}{1-\tan (\alpha+\beta)\tan {\beta}}+\frac{\tan (\alpha+\beta)}{1-\tan (\alpha+\beta)\tan {\beta}}=\tan (2\alpha +\beta)=\frac{AY}{AX}.</cmath> | <cmath>\frac{\tan \alpha}{1-\tan (\alpha+\beta)\tan {\beta}}+\frac{\tan (\alpha+\beta)}{1-\tan (\alpha+\beta)\tan {\beta}}=\tan (2\alpha +\beta)=\frac{AY}{AX}.</cmath> | ||

+ | |||

+ | ==Solution 4== | ||

+ | First, since <math>XY</math> is the diameter and <math>A</math>, <math>B</math>, and <math>C</math> lie on the circle, <cmath>\angle {XAY} = \angle {XBY} = \angle{XCY} = 90</cmath>. Next, because <math>AZ</math> and <math>CY</math> are both perpendicular to <math>CX</math>, we have <math>AZ</math> to be parallel to <math>CY</math>. | ||

+ | |||

+ | Now looking at quadrilateral <math>APZX</math>, we see that this is cyclic because <cmath>\angle {PAX} + \angle {PZX} = 90+90 = 180.</cmath> Set <math>\alpha = \angle{BXA} = \angle{BYA}</math>, and <math>\beta = \angle{BXC} = \angle{BYC}</math>. | ||

+ | Now, <cmath>\angle{AYC} = \angle{YAZ}</cmath> since <math>AZ</math> and <math>CY</math> are parallel. | ||

+ | Also, <cmath>\angle{PAZ} = \angle{PXZ} = \alpha + \beta.</cmath> | ||

+ | That means <cmath>\angle{PXZ} = \angle{PXQ} + \angle{QXZ} = \beta + \angle{QXZ} = \alpha + \beta</cmath> so <cmath>\angle{QXZ} = \alpha.</cmath> This means <math>\angle{QXZ} = \angle{YBC} = \alpha</math>, so <math>BC</math> and <math>AY</math> are parallel. | ||

+ | Finally, we can look at the equation. | ||

+ | We know <cmath>XP\cos{\alpha} = AX,</cmath> so <math>XP = \frac{AX}{\cos{\alpha}}.</math> | ||

+ | We also know <cmath>XQ\cos(\alpha+\beta) = AX,</cmath> so <math>XQ = \frac{AX}{\cos(\alpha+\beta)}.</math> | ||

+ | Plugging this into the LHS of the equation, we get <cmath>\frac{BY\cos\alpha}{AX}+\frac{CY\cos(\alpha+\beta)}{AZ}.</cmath> | ||

+ | Now, let <math>H_1</math> and <math>H_2</math> be the points on <math>AY</math> such that <math>BH_1</math> and <math>CH_2</math> are perpendicular to <math>AY</math>. Also, since <math>\angle{AYB} = \angle{CXY}</math>, their arcs have equal length, and <math>AB=CY</math>. | ||

+ | Now, the LHS is simplified even more to <cmath>\frac{AB\cos\alpha}{AX}+\frac{CY\cos(\alpha+\beta)}{AZ}</cmath> which is equal to <cmath>\frac{AH_1+YH_1}{AZ}</cmath> which is equal to <cmath>\frac{AY}{AX}.</cmath> This completes the proof. | ||

+ | |||

+ | ~jeteagle | ||

{{MAA Notice}} | {{MAA Notice}} |

## Revision as of 11:10, 24 July 2019

## Problem

Quadrilateral is inscribed in the semicircle with diameter . Segments and meet at . Point is the foot of the perpendicular from to line . Point lies on such that line is perpendicular to line . Let be the intersection of segments and . Prove that

## Solution 1

Let us use coordinates. Let O, the center of the circle, be (0,0). WLOG the radius of the circle is 1, so set Y (1,0) and X (-1,0). Also, for arbitrary constants and set A and B . Now, let's use our coordinate tools. It is easily derived that the equation of is and the equation of is , where and are defined appropriately. Thus, by equating the y's in the equation we find the intersection of these lines, , is . Also, . It shall be left to the reader to find the slope of , the coordinates of Q and C, and use the distance formula to verify that .

## Solution 2

First of all

since the quadrilateral is cyclic, and triangle is rectangle, and is orthogonal to . Now

because is cyclic and we have proved that

so is parallel to , and Now by Ptolomey's theorem on we have we see that triangles and are similar since and is already proven, so Substituting yields dividing by We get Now triangles , and are similar so but also triangles and are similar and we get Comparing we have, Substituting, Dividing the new relation by and multiplying by we get but since triangles and are similar, because and since Substituting again we get Now since triangles and are similar we have and by the similarity of and , we get so substituting, and separating terms we get In the beginning we prove that and so

## Solution 3

It is obvious that for some value . Also, note that . Set We have and This gives Similarly, we can deduce that Adding gives

## Solution 4

First, since is the diameter and , , and lie on the circle, . Next, because and are both perpendicular to , we have to be parallel to .

Now looking at quadrilateral , we see that this is cyclic because Set , and . Now, since and are parallel. Also, That means so This means , so and are parallel. Finally, we can look at the equation. We know so We also know so Plugging this into the LHS of the equation, we get Now, let and be the points on such that and are perpendicular to . Also, since , their arcs have equal length, and . Now, the LHS is simplified even more to which is equal to which is equal to This completes the proof.

~jeteagle

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