Difference between revisions of "2024 USAJMO Problems/Problem 1"
Technodoggo (talk | contribs) (sol 1 womp womp) |
(→Solution 3) |
||
(9 intermediate revisions by 5 users not shown) | |||
Line 3: | Line 3: | ||
==Problem== | ==Problem== | ||
− | + | 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. | |
==Solution 1== | ==Solution 1== | ||
Line 9: | Line 9: | ||
First, let <math>E</math> and <math>F</math> be the midpoints of <math>AB</math> and <math>CD</math>, respectively. It is clear that <math>AE=BE=3.5</math>, <math>PE=QE=0.5</math>, <math>DF=CF=4</math>, and <math>SF=RF=2</math>. Also, let <math>O</math> be the circumcenter of <math>ABCD</math>. | First, let <math>E</math> and <math>F</math> be the midpoints of <math>AB</math> and <math>CD</math>, respectively. It is clear that <math>AE=BE=3.5</math>, <math>PE=QE=0.5</math>, <math>DF=CF=4</math>, and <math>SF=RF=2</math>. Also, let <math>O</math> be the circumcenter of <math>ABCD</math>. | ||
− | + | <asy> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ | |
import graph; size(12cm); | import graph; size(12cm); | ||
real labelscalefactor = 0.5; /* changes label-to-point distance */ | real labelscalefactor = 0.5; /* changes label-to-point distance */ | ||
Line 26: | Line 26: | ||
/* dots and labels */ | /* dots and labels */ | ||
dot((2.92,-3.28),dotstyle); | dot((2.92,-3.28),dotstyle); | ||
− | label(" | + | label("$O$", (2.43,-3.56), NE * labelscalefactor); |
dot((-2.52,-1.01),dotstyle); | dot((-2.52,-1.01),dotstyle); | ||
− | label(" | + | label("$A$", (-2.91,-0.91), NE * labelscalefactor); |
dot((3.46,2.59),linewidth(4pt) + dotstyle); | dot((3.46,2.59),linewidth(4pt) + dotstyle); | ||
− | label(" | + | label("$B$", (3.49,2.78), NE * labelscalefactor); |
dot((7.59,-6.88),dotstyle); | dot((7.59,-6.88),dotstyle); | ||
− | label(" | + | label("$C$", (7.82,-7.24), NE * labelscalefactor); |
dot((-0.29,-8.22),linewidth(4pt) + dotstyle); | dot((-0.29,-8.22),linewidth(4pt) + dotstyle); | ||
− | label(" | + | label("$D$", (-0.53,-8.62), NE * labelscalefactor); |
dot((0.03,0.52),linewidth(4pt) + dotstyle); | dot((0.03,0.52),linewidth(4pt) + dotstyle); | ||
− | label(" | + | label("$P$", (-0.13,0.67), NE * labelscalefactor); |
dot((0.89,1.04),linewidth(4pt) + dotstyle); | dot((0.89,1.04),linewidth(4pt) + dotstyle); | ||
− | label(" | + | label("$Q$", (0.62,1.16), NE * labelscalefactor); |
dot((5.61,-7.22),linewidth(4pt) + dotstyle); | dot((5.61,-7.22),linewidth(4pt) + dotstyle); | ||
− | label(" | + | label("$R$", (5.70,-7.05), NE * labelscalefactor); |
dot((1.67,-7.89),linewidth(4pt) + dotstyle); | dot((1.67,-7.89),linewidth(4pt) + dotstyle); | ||
− | label(" | + | label("$S$", (1.75,-7.73), NE * labelscalefactor); |
dot((0.46,0.78),linewidth(4pt) + dotstyle); | dot((0.46,0.78),linewidth(4pt) + dotstyle); | ||
− | label(" | + | label("$E$", (0.26,0.93), NE * labelscalefactor); |
dot((3.64,-7.55),linewidth(4pt) + dotstyle); | dot((3.64,-7.55),linewidth(4pt) + dotstyle); | ||
− | label(" | + | label("$F$", (3.73,-7.39), NE * labelscalefactor); |
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | ||
− | /* end of picture */ | + | /* end of picture */</asy> |
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 <math>OE\perp AB</math> and <math>OF\perp CD</math>. Since <math>E</math> and <math>F</math> are also bisectors of <math>PQ</math> and <math>RS</math>, respectively, if <math>PQRS</math> is indeed a cyclic quadrilateral, then its circumcenter is also at <math>O</math>. Thus, it suffices to show that <math>OP=OQ=OR=OS</math>. | 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 <math>OE\perp AB</math> and <math>OF\perp CD</math>. Since <math>E</math> and <math>F</math> are also bisectors of <math>PQ</math> and <math>RS</math>, respectively, if <math>PQRS</math> is indeed a cyclic quadrilateral, then its circumcenter is also at <math>O</math>. Thus, it suffices to show that <math>OP=OQ=OR=OS</math>. | ||
Line 56: | Line 56: | ||
Draw the segments connecting <math>O</math> to <math>B</math>, <math>Q</math>, <math>C</math>, and <math>R</math>. | Draw the segments connecting <math>O</math> to <math>B</math>, <math>Q</math>, <math>C</math>, and <math>R</math>. | ||
− | + | <asy> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ | |
import graph; size(12cm); | import graph; size(12cm); | ||
real labelscalefactor = 0.5; /* changes label-to-point distance */ | real labelscalefactor = 0.5; /* changes label-to-point distance */ | ||
Line 79: | Line 79: | ||
/* dots and labels */ | /* dots and labels */ | ||
dot((2.92,-3.28),dotstyle); | dot((2.92,-3.28),dotstyle); | ||
− | label(" | + | label("$O$", (2.43,-3.56), NE * labelscalefactor); |
dot((-2.52,-1.01),dotstyle); | dot((-2.52,-1.01),dotstyle); | ||
− | label(" | + | label("$A$", (-2.91,-0.91), NE * labelscalefactor); |
dot((3.46,2.59),linewidth(1pt) + dotstyle); | dot((3.46,2.59),linewidth(1pt) + dotstyle); | ||
− | label(" | + | label("$B$", (3.49,2.78), NE * labelscalefactor); |
dot((7.59,-6.88),dotstyle); | dot((7.59,-6.88),dotstyle); | ||
− | label(" | + | label("$C$", (7.82,-7.24), NE * labelscalefactor); |
dot((-0.29,-8.22),linewidth(1pt) + dotstyle); | dot((-0.29,-8.22),linewidth(1pt) + dotstyle); | ||
− | label(" | + | label("$D$", (-0.53,-8.62), NE * labelscalefactor); |
dot((0.03,0.52),linewidth(1pt) + dotstyle); | dot((0.03,0.52),linewidth(1pt) + dotstyle); | ||
− | label(" | + | label("$P$", (-0.13,0.67), NE * labelscalefactor); |
dot((0.89,1.04),linewidth(1pt) + dotstyle); | dot((0.89,1.04),linewidth(1pt) + dotstyle); | ||
− | label(" | + | label("$Q$", (0.62,1.16), NE * labelscalefactor); |
dot((5.61,-7.22),linewidth(1pt) + dotstyle); | dot((5.61,-7.22),linewidth(1pt) + dotstyle); | ||
− | label(" | + | label("$R$", (5.70,-7.05), NE * labelscalefactor); |
dot((1.67,-7.89),linewidth(1pt) + dotstyle); | dot((1.67,-7.89),linewidth(1pt) + dotstyle); | ||
− | label(" | + | label("$S$", (1.75,-7.73), NE * labelscalefactor); |
dot((0.46,0.78),linewidth(1pt) + dotstyle); | dot((0.46,0.78),linewidth(1pt) + dotstyle); | ||
− | label(" | + | label("$E$", (0.26,0.93), NE * labelscalefactor); |
dot((3.64,-7.55),linewidth(1pt) + dotstyle); | dot((3.64,-7.55),linewidth(1pt) + dotstyle); | ||
− | label(" | + | label("$F$", (3.73,-7.39), NE * labelscalefactor); |
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | ||
− | /* end of picture */ | + | /* end of picture */</asy> |
Also, let <math>r</math> be the circumradius of <math>ABCD</math>. This means that <math>AO=BO=CO=DO=r</math>. Recall that <math>\angle BEO=90^\circ</math> and <math>\angle CFO=90^\circ</math>. Notice the several right triangles in our figure. | Also, let <math>r</math> be the circumradius of <math>ABCD</math>. This means that <math>AO=BO=CO=DO=r</math>. Recall that <math>\angle BEO=90^\circ</math> and <math>\angle CFO=90^\circ</math>. Notice the several right triangles in our figure. | ||
Line 116: | Line 116: | ||
~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)== | ||
+ | [Will add when have time] | ||
+ | ~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}} |
Revision as of 09:38, 24 April 2024
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.
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)
[Will add when have time] ~KevinChen_Yay
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.