Difference between revisions of "2013 USAJMO Problems/Problem 5"
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− | + | First of all <math> \angle BXY = \angle PAZ =\angle AXQ =\angle AXC</math>, 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 <math>\angle BXY =\angle BAY =\angle AXC</math> because <math>XABY</math> is cyclic and we have proved that <math>\angle AXC = \angle BXY</math>, so <math>BC</math> is parallel to <math>AY</math>, and <math>AC=BY</math>, <math>CY=AB</math>. Now by Ptolomey's theorem on <math>APZX</math>, we have <math>(AX)(PZ)+(AP)(XZ)=(AZ)(PX)</math>, we see that triangles <math>PXZ</math> and <math>QXA</math> are similar since <math>\angle QAX= \angle PZX= 90</math> and <math>\angle AXC = \angle BXY</math>, already proven, so <math>(AX)(PZ)=(AQ)(XZ)</math>, substituting we get <math>(AQ)(XZ)+(AP)(XZ)=(AZ)(PX)</math>, dividing by <math>(PX)(XZ)</math>, we get <math>\frac {AQ+AP}{XP} = \frac {AZ}{XZ}</math>. Now triangles <math>AYZ</math>, and <math>XYP</math> are similar so <math>\frac {AY}{AZ}= \frac {XY}{XP}</math>, but also triangles <math>XPY</math> and <math>XZB</math> are similar and we get <math>\frac {XY}{XP}= \frac {XB}{XZ}</math>, comparing we have, <math>\frac {AY}{XB}= \frac {AZ}{XZ}</math> substituting, <math>\frac {AQ+AP}{XP}= \frac {AY}{XB}</math>. Dividing the new relation by <math>AX</math> and multiplying by <math>XB</math> we get <math>\frac{XB(AQ+AP)}{(XP)(AX)} = \frac {AY}{AX}</math>, but <math>\frac {XB}{AX}= \frac {XY}{XQ}</math>, since triangles <math>AXB</math> and <math>QXY</math> are similar, because <math>\angle AYX= \angle ABX</math> and <math>\angle AXB= \angle CXY</math> since <math>CY=AB</math>. Substituting again we get <math>\frac {XY(AQ)+XY(AP)}{(XP)(XQ)} =\frac {AY}{AX}</math>. Now since triangles <math>ACQ</math> and <math>XYQ</math> are similar we have <math>XY(AQ)=AC(XQ)</math> and by the similarity of <math>APB</math> and <math>XPY</math>, we get <math>AB(CP)=XY(AP)</math> so substituting, and separating terms we get <math>\frac{AC}{XP} + \frac{AB}{XQ} = \frac{AY}{AX}</math>, but in the beginning we prove that <math>AC=BY</math> and <math>AB=CY</math> so <math>\frac{BY}{XP} + \frac{CY}{XQ} = \frac{AY}{AX}</math>, and we are done. | |
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Revision as of 16:21, 2 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
, already proven, so
, substituting we get
, 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
, but in the beginning we prove that
and
so
, and we are done.
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