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

m (→Solution 1) |
(→Solution) |
||

(13 intermediate revisions by 4 users not shown) | |||

Line 2: | Line 2: | ||

Quadrilateral <math>XABY</math> is inscribed in the semicircle <math>\omega</math> with diameter <math>XY</math>. Segments <math>AY</math> and <math>BX</math> meet at <math>P</math>. Point <math>Z</math> is the foot of the perpendicular from <math>P</math> to line <math>XY</math>. Point <math>C</math> lies on <math>\omega</math> such that line <math>XC</math> is perpendicular to line <math>AZ</math>. Let <math>Q</math> be the intersection of segments <math>AY</math> and <math>XC</math>. Prove that <cmath>\dfrac{BY}{XP}+\dfrac{CY}{XQ}=\dfrac{AY}{AX}.</cmath> | Quadrilateral <math>XABY</math> is inscribed in the semicircle <math>\omega</math> with diameter <math>XY</math>. Segments <math>AY</math> and <math>BX</math> meet at <math>P</math>. Point <math>Z</math> is the foot of the perpendicular from <math>P</math> to line <math>XY</math>. Point <math>C</math> lies on <math>\omega</math> such that line <math>XC</math> is perpendicular to line <math>AZ</math>. Let <math>Q</math> be the intersection of segments <math>AY</math> and <math>XC</math>. Prove that <cmath>\dfrac{BY}{XP}+\dfrac{CY}{XQ}=\dfrac{AY}{AX}.</cmath> | ||

+ | |||

+ | ==Solution== | ||

+ | Using the Law of Sines and simplifying, we have <cmath>\frac{BY}{XP}+\frac{CY}{XQ}=\frac{\sin \angle PXY \sin \angle XPY+\sin \angle QXY\sin \angle XQY}{\sin \angle AYX}.</cmath> | ||

+ | |||

+ | It is easy to see that <math>APZX</math> is cyclic. Also, we are given <math>XQ\perp AZ</math>. Then we have | ||

+ | <cmath>\begin{align*} | ||

+ | \frac{\sin \angle PXY\sin \angle XPY+\sin \angle QXY\sin \angle XQY}{\sin \angle AYX} &= \frac{\sin \angle YAZ\sin \angle XZA+\cos \angle XZA\cos \angle YAZ}{\sin \angle AYX} \\ | ||

+ | &= \frac{\cos(\angle XZA-\angle YAZ)}{\sin \angle AYX} \\ | ||

+ | &= \frac{\cos \angle AYX}{\sin \angle AYX} \\ | ||

+ | &= \cot AYX \\ | ||

+ | &= \frac{AY}{AX}, | ||

+ | \end{align*}</cmath> and we are done. | ||

==Solution 1== | ==Solution 1== | ||

Line 7: | Line 19: | ||

==Solution 2== | ==Solution 2== | ||

− | First < | + | First of all |

+ | |||

+ | <cmath>\angle BXY = \angle PAZ =\angle AXQ =\angle AXC</cmath> since the quadrilateral <math>APZX</math> is cyclic, and triangle <math>AXQ</math> is rectangle, and <math>CX</math> is orthogonal to <math>AZ</math>. Now | ||

+ | |||

+ | <cmath>\angle BXY =\angle BAY =\angle AXC</cmath> because <math>XABY</math> is cyclic and we have proved that | ||

+ | |||

+ | <cmath>\angle AXC = \angle BXY</cmath> so <math>BC</math> is parallel to <math>AY</math>, and <cmath>AC=BY, CY=AB</cmath> Now by Ptolomey's theorem on <math>APZX</math> we have <cmath>(AX)(PZ)+(AP)(XZ)=(AZ)(PX)</cmath> we see that triangles <math>PXZ</math> and <math>QXA</math> are similar since <cmath>\angle QAX= \angle PZX= 90</cmath> and <cmath>\angle AXC = \angle BXY</cmath> is already proven, so <cmath>(AX)(PZ)=(AQ)(XZ)</cmath> Substituting yields <cmath>(AQ)(XZ)+(AP)(XZ)=(AZ)(PX)</cmath> dividing by <math>(PX)(XZ)</math> We get <cmath>\frac {AQ+AP}{XP} = \frac {AZ}{XZ}</cmath> Now triangles <math>AYZ</math>, and <math>XYP</math> are similar so <cmath>\frac {AY}{AZ}= \frac {XY}{XP}</cmath> but also triangles <math>XPY</math> and <math>XZB</math> are similar and we get <cmath>\frac {XY}{XP}= \frac {XB}{XZ}</cmath> Comparing we have, <cmath>\frac {AY}{XB}= \frac {AZ}{XZ}</cmath> Substituting, <cmath>\frac {AQ+AP}{XP}= \frac {AY}{XB}</cmath> Dividing the new relation by <math>AX</math> and multiplying by <math>XB</math> we get <cmath>\frac{XB(AQ+AP)}{(XP)(AX)} = \frac {AY}{AX}</cmath> but <cmath>\frac {XB}{AX}= \frac {XY}{XQ}</cmath> since triangles <math>AXB</math> and <math>QXY</math> are similar, because <cmath>\angle AYX= \angle ABX</cmath> and <cmath>\angle AXB= \angle CXY</cmath> since <math>CY=AB</math> Substituting again we get <cmath>\frac {XY(AQ)+XY(AP)}{(XP)(XQ)} =\frac {AY}{AX}</cmath> Now since triangles <math>ACQ</math> and <math>XYQ</math> are similar we have <cmath>XY(AQ)=AC(XQ)</cmath> and by the similarity of <math>APB</math> and <math>XPY</math>, we get <cmath>AB(CP)=XY(AP)</cmath> so substituting, and separating terms we get <cmath>\frac{AC}{XP} + \frac{AB}{XQ} = \frac{AY}{AX}</cmath> In the beginning we prove that <math>AC=BY</math> and <math>AB=CY</math> so <cmath>\frac{BY}{XP} + \frac{CY}{XQ} = \frac{AY}{AX}</cmath> | ||

+ | <math>\blacksquare</math> | ||

+ | |||

+ | ==Solution 3== | ||

+ | It is obvious that | ||

+ | <cmath>\angle AXB=\angle CXY=\alpha</cmath> | ||

+ | for some value <math>\alpha</math>. Also, note that <math>\angle BYA=\alpha</math>. Set | ||

+ | <cmath>\angle BXC=\angle BYC=\beta.</cmath> | ||

+ | We have | ||

+ | <cmath>\frac{XC}{CY}=\tan {\angle CYZ}=\tan (90-\alpha)</cmath> | ||

+ | and | ||

+ | <cmath>\frac{CQ}{CY}=\tan {\angle CYQ}=\tan (\alpha+\beta).</cmath> | ||

+ | This gives | ||

+ | <cmath>\frac{CY}{XQ}=\frac{1}{\tan (90-\alpha)-\tan (\alpha+\beta)}.</cmath> | ||

+ | Similarly, we can deduce that | ||

+ | <cmath>\frac{BY}{XP}=\frac{1}{\tan (90-\alpha-\beta)-\tan {\alpha}}.</cmath> | ||

+ | Adding gives | ||

+ | <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</math> be the point on <math>AY</math> such that <math>BH</math> is 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+YH}{AZ}</cmath> which is equal to <cmath>\frac{AY}{AX}.</cmath> This completes the proof. | ||

+ | ~jeteagle | ||

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

## Latest revision as of 18:05, 1 October 2021

## 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

Using the Law of Sines and simplifying, we have

It is easy to see that is cyclic. Also, we are given . Then we have and we are done.

## 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 be the point on such that is 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

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