Difference between revisions of "2013 USAJMO Problems/Problem 5"
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+ | <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> | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:49, 25 April 2015
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
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