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==Solution 2== | ==Solution 2== | ||
− | We can consider two cases: <math>AB \parallel CD</math> or <math>AB \nparallel CD.</math> The first case is trivial, as <math>PQ \parallel RS</math> and we are done. For the second case, WLOG, assume that <math>A</math> and <math>C</math> are located on <math>XB</math> and <math>XD</math> respectively. Extend <math>AB</math> and <math>CD</math> to a point <math>X,</math> and by Power of a Point, we have <cmath>XA\cdot XB = XC \cdot XD,</cmath> which may be written as <cmath>XA \cdot (XA+7) = XC \cdot (XC+8),</cmath> or <cmath>XA^2 + 7XA = XC^2 + 8XC.</cmath> We can translate this to <cmath>XA^2 + 7XA +12 = XC^2 + 8XC +12,</cmath> so <cmath>XP\cdot XQ = (XA+3)(XA+4)=(XC+2)(XC+6)= XR\cdot XS,</cmath> and therefore by the Converse of Power of a Point <math>PQRS</math> is cyclic, and we are done. | + | We can consider two cases: <math>AB \parallel CD</math> or <math>AB \nparallel CD.</math> The first case is trivial, as <math>PQ \parallel RS</math> and we are done due to symmetry. For the second case, WLOG, assume that <math>A</math> and <math>C</math> are located on <math>XB</math> and <math>XD</math> respectively. Extend <math>AB</math> and <math>CD</math> to a point <math>X,</math> and by Power of a Point, we have <cmath>XA\cdot XB = XC \cdot XD,</cmath> which may be written as <cmath>XA \cdot (XA+7) = XC \cdot (XC+8),</cmath> or <cmath>XA^2 + 7XA = XC^2 + 8XC.</cmath> We can translate this to <cmath>XA^2 + 7XA +12 = XC^2 + 8XC +12,</cmath> so <cmath>XP\cdot XQ = (XA+3)(XA+4)=(XC+2)(XC+6)= XR\cdot XS,</cmath> and therefore by the Converse of Power of a Point <math>PQRS</math> is cyclic, and we are done. |
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Revision as of 18:46, 24 March 2024
Contents
[hide]Problem
Let be a cyclic quadrilateral with
and
. Points
and
are selected on segment
such that
. Points
and
are selected on segment
such that
. Prove that
is a cyclic quadrilateral.
Solution 1
First, let and
be the midpoints of
and
, respectively. It is clear that
,
,
, and
. Also, let
be the circumcenter of
.
By properties of cyclic quadrilaterals, we know that the circumcenter of a cyclic quadrilateral is the intersection of its sides' perpendicular bisectors. This implies that and
. Since
and
are also bisectors of
and
, respectively, if
is indeed a cyclic quadrilateral, then its circumcenter is also at
. Thus, it suffices to show that
.
Notice that ,
, and
. By SAS congruency,
. Similarly, we find that
and
. We now need only to show that these two pairs are equal to each other.
Draw the segments connecting to
,
,
, and
.
Also, let be the circumradius of
. This means that
. Recall that
and
. Notice the several right triangles in our figure.
Let us apply Pythagorean Theorem on . We can see that
Let us again apply Pythagorean Theorem on . We can see that
Let us apply Pythagorean Theorem on . We get
.
We finally apply Pythagorean Theorem on . This becomes
.
This is the same expression as we got for . Thus,
, and recalling that
and
, we have shown that
. We are done. QED
~Technodoggo
Solution 2
We can consider two cases: or
The first case is trivial, as
and we are done due to symmetry. For the second case, WLOG, assume that
and
are located on
and
respectively. Extend
and
to a point
and by Power of a Point, we have
which may be written as
or
We can translate this to
so
and therefore by the Converse of Power of a Point
is cyclic, and we are done.
See Also
2024 USAJMO (Problems • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.