Difference between revisions of "2013 USAJMO Problems/Problem 5"
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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== | ||
− | 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 <math>a</math> and <math>b</math> set A <math>(\cos a, \sin a)</math> and B <math>(\cos b, \sin b)</math>. Now, let's use our coordinate tools. It is easily derived that the equation of <math>BX</math> is <math>y = \frac{\sin b}{1 + \cos b}(x + 1) = v(x+1)</math> and the equation of <math>AY</math> is <math>y = \frac{\sin a}{1 - \cos a}(x - 1) = u(x-1)</math>, where <math>u</math> and <math>v</math> are defined appropriately. Thus, by equating the y's in the equation we find the intersection of these lines, <math>P</math>, is <math>(\frac{u-v}{u+v}, \frac{2uv}{u+v})</math>. Also, <math>Z(\frac{u-v}{u+v}, 0)</math>. It shall be left to the reader to find the slope of <math>AZ</math>, the coordinates of Q and C, and use the distance formula to verify that <math>\frac{BY}{XP} + \frac{CY}{XQ} = \frac{AY}{AX}</math>. | + | 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 <math>a</math> and <math>b</math> set A <math>(\cos a, \sin a)</math> and B <math>(\cos b, \sin b)</math>. Now, let's use our coordinate tools. It is easily derived that the equation of <math>BX</math> is <math>y = \frac{\sin b}{1 + \cos b}(x + 1) = v(x+1)</math> and the equation of <math>AY</math> is <math>y = \frac{\sin a}{1 - \cos a}(x - 1) = u(x-1)</math>, where <math>u</math> and <math>v</math> are defined appropriately. Thus, by equating the y's in the equation we find the intersection of these lines, <math>P</math>, is <math>\left(\frac{u-v}{u+v}\right), \frac{2uv}{u+v})</math>. Also, <math>Z\left(\frac{u-v}{u+v}\right), 0)</math>. It shall be left to the reader to find the slope of <math>AZ</math>, the coordinates of Q and C, and use the distance formula to verify that <math>\frac{BY}{XP} + \frac{CY}{XQ} = \frac{AY}{AX}</math>. |
==Solution 2== | ==Solution 2== | ||
− | + | 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.