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Let <math>ABCD</math> be a cyclic quadrilateral with <math>AB = 7</math> and <math>CD = 8</math>. Points <math>P</math> and <math>Q</math> are selected on segment <math>AB</math> such that <math>AP = BQ = 3</math>. Points <math>R</math> and <math>S</math> are selected on segment <math>CD</math> such that <math>CR = DS = 2</math>. Prove that <math>PQRS</math> is a cyclic quadrilateral. | Let <math>ABCD</math> be a cyclic quadrilateral with <math>AB = 7</math> and <math>CD = 8</math>. Points <math>P</math> and <math>Q</math> are selected on segment <math>AB</math> such that <math>AP = BQ = 3</math>. Points <math>R</math> and <math>S</math> are selected on segment <math>CD</math> such that <math>CR = DS = 2</math>. Prove that <math>PQRS</math> is a cyclic quadrilateral. | ||
+ | |||
+ | ==Video Solution 1 by MegaMath - No Casework Needed!== | ||
+ | |||
+ | https://www.youtube.com/watch?v=5sZQrCHiaqY | ||
==Solution 1== | ==Solution 1== | ||
Line 116: | Line 120: | ||
~Technodoggo | ~Technodoggo | ||
+ | |||
+ | ==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 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. | ||
+ | |||
+ | - [https://artofproblemsolving.com/wiki/index.php/User:Spectraldragon8 spectraldragon8] | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | All 4 corners of <math>PQRS</math> have equal power of a point (<math>12</math>) with respect to the circle <math>(ABCD)</math>, with center <math>O</math>. | ||
+ | |||
+ | Draw diameters (of length <math>AQ</math>) of circle <math>(ABCD)</math> through <math>Q</math> and <math>S</math>, with length <math>A</math>. Let <math>q</math> be the distance from <math>Q</math> to the circle along a diameter, and likewise <math>s</math> be distance from <math>S</math> to the circle. | ||
+ | |||
+ | Then <math>q(AQ-q) = s(AQ-s) = 12</math> and <math>q,s < AQ/2</math> (radius). Therefore, <math>q=s</math> and <math>AQ/2 -q = AQ/2 -s</math>. But <math>AQ-q=OQ</math>, <math>AQ-s=OS</math>, <math>OQ = OP</math> and <math>OS = OR</math> by symmetry around the perpendicular bisectors of <math>PQ</math> and <math>RS</math>, so <math>P,Q,R,S</math> are all equidistant from <math>O</math>, forming a circumcircle around <math>PQRS</math>. | ||
+ | |||
+ | -BraveCobra22aops and oinava | ||
+ | |||
+ | |||
+ | ==Solution 4 (Coord Bash)== | ||
+ | |||
+ | Let <math>A(2a_1,2a_2)</math>, <math>B(2b_1,2b_2)</math>, <math>C(2c_1,2c_2)</math>, <math>D(2d_1,2d_2)</math>, and the circumcenter of quadrilateral <math>ABCD</math> be <math>O(0,0)</math>. | ||
+ | |||
+ | Let's list what we know from the givens: | ||
+ | Since the radii of a circle are equal in length, we can let <cmath>k=a_1^2+a_2^2=b_1^2+b_2^2=c_1^2+c_2^2=d_1^2+d_2^2.</cmath> | ||
+ | |||
+ | From the distance formula on <math>AB</math> and <math>CD</math>, we can simplify and get <cmath>a_1^2+a_2^2+b_1^2+b_2^2-2(a_1b_1+a_2b_2)=\frac{49}{4} \Rightarrow 2(a_1b_1+a_2b_2)=2k-\frac{49}{4}</cmath> and <cmath>c_1^2+c_2^2+d_1^2+d_2^2-2(c_1d_1+c_2d_2)=16 \Rightarrow 2(c_1d_1+c_2d_2)=2k-16</cmath> using the above substitution with <math>k</math>. | ||
+ | |||
+ | Since <math>P</math>, <math>Q</math>, <math>R</math>, and <math>S</math> are weighted points on <math>AB</math> and <math>CD</math>, we can get <cmath>P\left(\frac{8a_1+6b_1}{7},\frac{8a_2+6b_2}{7}\right) \ \text{and} \ Q\left(\frac{6a_1+8b_1}{7},\frac{6a_2+8b_2}{7}\right)</cmath> along with <cmath>R\left(\frac{3c_1+d_1}{2},\frac{3c_2+d_2}{2}\right) \ \text{and} \ S\left(\frac{c_1+3d_1}{2},\frac{c_1+3d_1}{2}\right).</cmath> | ||
+ | |||
+ | Now, we want to show that the perpendicular bisectors of the sides of quadrilateral <math>PQRS</math> are concurrent to prove that it is cyclic. Moreover, we suspect that the circumcenter of quadrilateral <math>PQRS</math> is the circumcenter of quadrilateral <math>ABCD</math>, both of them being <math>(0,0)</math>. | ||
+ | |||
+ | We only list the linear equations of the perpendicular bisectors of lines <math>PS</math> and <math>QR</math>, as the perpendicular bisectors of lines <math>PQ</math> and <math>RS</math> are the same as the perpendicular bisectors of lines <math>AB</math> and <math>CD</math>, respectively. | ||
+ | |||
+ | First, we analyze the perpendicular bisector of line <math>PS</math>. Notice that it has slope <cmath>-\frac{16a_1+12b_1-7c_1-21d_1}{16a_2+12b_2-7c_2-21d_2}</cmath> and point <cmath>\left(\frac{16a_1+12b_1+7c_1+21d_1}{28}, \frac{16a_2+12b_2+7c_2+21d_2}{28}\right).</cmath> The point-slope form equation would thus be <cmath>y-\frac{16a_2+12b_2+7c_2+21d_2}{28}=-\frac{16a_1+12b_1-7c_1-21d_1}{16a_2+12b_2-7c_2-21d_2}\left(x-\frac{16a_1+12b_1+7c_1+21d_1}{28}\right).</cmath> Since we claim that <math>(0,0)</math> is on the line, we substitute to get <cmath>(16a_1+12b_1+7c_1+21d_1)(16a_1+12b_1-7c_1-21d_1)=-(16a_2+12b_2+7c_2+21d_2)(16a_2+12b_2-7c_2-21d_2),</cmath> and this simplifies to <cmath>16\left(16(a_1^2+a_2^2)+24(a_1b_1+a_2b_2)+9(b_1^2+b_2^2)\right)=49\left((c_1^2+c_2^2)+6(c_1d_1+c_2d_2)+9(d_1^2+d_2^2)\right).</cmath> Using substitutions from the first three equations, this becomes <cmath>16\left(16k+12\left(2k-\frac{49}{4}\right)+9k\right)=49\left(k+3(2k-16)+9k\right)</cmath> <cmath>\Rightarrow 16(49k-3\cdot49)=49(16k-3\cdot16)</cmath> <cmath>\Rightarrow k-3=k-3,</cmath> which is true, implying that <math>(0,0)</math> does indeed satisfy the equation. | ||
+ | |||
+ | Analogously, we can show that the perpendicular bisector of line <math>QR</math> also passes through the origin. | ||
+ | |||
+ | Since the perpendicular bisectors of the sides of quadrilateral <math>PQRS</math> all intersect at the same point, namely <math>(0,0)</math>, which is also the circumcenter of quadrilateral <math>ABCD</math>, we can conclude that <math>PQRS</math> is a cyclic quadrilateral. | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Kevinchen_yay KevinChen_Yay] | ||
==See Also== | ==See Also== | ||
{{USAJMO newbox|year=2024|before=First Question|num-a=2}} | {{USAJMO newbox|year=2024|before=First Question|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 15:42, 30 April 2024
Contents
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.
Video Solution 1 by MegaMath - No Casework Needed!
https://www.youtube.com/watch?v=5sZQrCHiaqY
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.
Solution 3
All 4 corners of have equal power of a point (
) with respect to the circle
, with center
.
Draw diameters (of length ) of circle
through
and
, with length
. Let
be the distance from
to the circle along a diameter, and likewise
be distance from
to the circle.
Then and
(radius). Therefore,
and
. But
,
,
and
by symmetry around the perpendicular bisectors of
and
, so
are all equidistant from
, forming a circumcircle around
.
-BraveCobra22aops and oinava
Solution 4 (Coord Bash)
Let ,
,
,
, and the circumcenter of quadrilateral
be
.
Let's list what we know from the givens:
Since the radii of a circle are equal in length, we can let
From the distance formula on and
, we can simplify and get
and
using the above substitution with
.
Since ,
,
, and
are weighted points on
and
, we can get
along with
Now, we want to show that the perpendicular bisectors of the sides of quadrilateral are concurrent to prove that it is cyclic. Moreover, we suspect that the circumcenter of quadrilateral
is the circumcenter of quadrilateral
, both of them being
.
We only list the linear equations of the perpendicular bisectors of lines and
, as the perpendicular bisectors of lines
and
are the same as the perpendicular bisectors of lines
and
, respectively.
First, we analyze the perpendicular bisector of line . Notice that it has slope
and point
The point-slope form equation would thus be
Since we claim that
is on the line, we substitute to get
and this simplifies to
Using substitutions from the first three equations, this becomes
which is true, implying that
does indeed satisfy the equation.
Analogously, we can show that the perpendicular bisector of line also passes through the origin.
Since the perpendicular bisectors of the sides of quadrilateral all intersect at the same point, namely
, which is also the circumcenter of quadrilateral
, we can conclude that
is a cyclic quadrilateral.
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.