2024 USAJMO Problems/Problem 1
Contents
[hide]Problem
sus
Solution 1
First, let and be the midpoints of and , respectively. It is clear that , , , and . Also, let be the circumcenter of .
[asy] import graph; size(12cm); real labelscalefactor = 0.5; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); pen dotstyle = black; real xmin = -12.19, xmax = 24.94, ymin = -15.45, ymax = 6.11; pen wrwrwr = rgb(0.38,0.38,0.38); draw(circle((2.92,-3.28), 5.90), linewidth(2) + wrwrwr); draw((-2.52,-1.01)--(3.46,2.59), linewidth(2) + wrwrwr); draw((7.59,-6.88)--(-0.29,-8.22), linewidth(2) + wrwrwr); draw((3.46,2.59)--(7.59,-6.88), linewidth(2) + wrwrwr); draw((-0.29,-8.22)--(-2.52,-1.01), linewidth(2) + wrwrwr); draw((0.03,0.52)--(1.67,-7.89), linewidth(2) + wrwrwr); draw((5.61,-7.22)--(0.89,1.04), linewidth(2) + wrwrwr); dot((2.92,-3.28),dotstyle); label("", (2.43,-3.56), NE * labelscalefactor); dot((-2.52,-1.01),dotstyle); label("", (-2.91,-0.91), NE * labelscalefactor); dot((3.46,2.59),linewidth(4pt) + dotstyle); label("", (3.49,2.78), NE * labelscalefactor); dot((7.59,-6.88),dotstyle); label("", (7.82,-7.24), NE * labelscalefactor); dot((-0.29,-8.22),linewidth(4pt) + dotstyle); label("", (-0.53,-8.62), NE * labelscalefactor); dot((0.03,0.52),linewidth(4pt) + dotstyle); label("", (-0.13,0.67), NE * labelscalefactor); dot((0.89,1.04),linewidth(4pt) + dotstyle); label("", (0.62,1.16), NE * labelscalefactor); dot((5.61,-7.22),linewidth(4pt) + dotstyle); label("", (5.70,-7.05), NE * labelscalefactor); dot((1.67,-7.89),linewidth(4pt) + dotstyle); label("", (1.75,-7.73), NE * labelscalefactor); dot((0.46,0.78),linewidth(4pt) + dotstyle); label("", (0.26,0.93), NE * labelscalefactor); dot((3.64,-7.55),linewidth(4pt) + dotstyle); label("", (3.73,-7.39), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/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 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 .
[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -12.19, xmax = 24.94, ymin = -15.45, ymax = 6.11; /* image dimensions */ pen wrwrwr = rgb(0.38,0.38,0.38);
/* draw figures */
draw(circle((2.92,-3.28), 5.90), linewidth(2) + wrwrwr); draw((-2.52,-1.01)--(3.46,2.59), linewidth(2) + wrwrwr); draw((7.59,-6.88)--(-0.29,-8.22), linewidth(2) + wrwrwr); draw((3.46,2.59)--(7.59,-6.88), linewidth(2) + wrwrwr); draw((-0.29,-8.22)--(-2.52,-1.01), linewidth(2) + wrwrwr); draw((0.03,0.52)--(1.67,-7.89), linewidth(2) + wrwrwr); draw((5.61,-7.22)--(0.89,1.04), linewidth(2) + wrwrwr); draw((0.46,0.78)--(2.92,-3.28), linewidth(2) + wrwrwr); draw((2.92,-3.28)--(3.64,-7.55), linewidth(2) + wrwrwr); draw((2.92,-3.28)--(7.59,-6.88), linewidth(2) + wrwrwr); draw((5.61,-7.22)--(2.92,-3.28), linewidth(2) + wrwrwr); draw((2.92,-3.28)--(3.46,2.59), linewidth(2) + wrwrwr); draw((2.92,-3.28)--(0.89,1.04), linewidth(2) + wrwrwr);
/* dots and labels */
dot((2.92,-3.28),dotstyle); label("", (2.43,-3.56), NE * labelscalefactor); dot((-2.52,-1.01),dotstyle); label("", (-2.91,-0.91), NE * labelscalefactor); dot((3.46,2.59),linewidth(1pt) + dotstyle); label("", (3.49,2.78), NE * labelscalefactor); dot((7.59,-6.88),dotstyle); label("", (7.82,-7.24), NE * labelscalefactor); dot((-0.29,-8.22),linewidth(1pt) + dotstyle); label("", (-0.53,-8.62), NE * labelscalefactor); dot((0.03,0.52),linewidth(1pt) + dotstyle); label("", (-0.13,0.67), NE * labelscalefactor); dot((0.89,1.04),linewidth(1pt) + dotstyle); label("", (0.62,1.16), NE * labelscalefactor); dot((5.61,-7.22),linewidth(1pt) + dotstyle); label("", (5.70,-7.05), NE * labelscalefactor); dot((1.67,-7.89),linewidth(1pt) + dotstyle); label("", (1.75,-7.73), NE * labelscalefactor); dot((0.46,0.78),linewidth(1pt) + dotstyle); label("", (0.26,0.93), NE * labelscalefactor); dot((3.64,-7.55),linewidth(1pt) + dotstyle); label("", (3.73,-7.39), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */[/asy]
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
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.