Difference between revisions of "2024 USAJMO Problems/Problem 1"
Technodoggo (talk | contribs) (sol 1 womp womp) |
Mathkiddie (talk | contribs) (→Solution 1) |
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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] | + | [asy] |
import graph; size(12cm); | import graph; size(12cm); | ||
− | real labelscalefactor = 0.5; | + | real labelscalefactor = 0.5; |
− | pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); | + | pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); |
− | pen dotstyle = black; | + | pen dotstyle = black; |
− | real xmin = -12.19, xmax = 24.94, ymin = -15.45, ymax = 6.11; | + | real xmin = -12.19, xmax = 24.94, ymin = -15.45, ymax = 6.11; |
pen wrwrwr = rgb(0.38,0.38,0.38); | pen wrwrwr = rgb(0.38,0.38,0.38); | ||
− | |||
draw(circle((2.92,-3.28), 5.90), linewidth(2) + wrwrwr); | draw(circle((2.92,-3.28), 5.90), linewidth(2) + wrwrwr); | ||
draw((-2.52,-1.01)--(3.46,2.59), linewidth(2) + wrwrwr); | draw((-2.52,-1.01)--(3.46,2.59), linewidth(2) + wrwrwr); | ||
Line 24: | Line 23: | ||
draw((0.03,0.52)--(1.67,-7.89), 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((5.61,-7.22)--(0.89,1.04), linewidth(2) + wrwrwr); | ||
− | |||
dot((2.92,-3.28),dotstyle); | dot((2.92,-3.28),dotstyle); | ||
label("<math>O</math>", (2.43,-3.56), NE * labelscalefactor); | label("<math>O</math>", (2.43,-3.56), NE * labelscalefactor); | ||
Line 48: | Line 46: | ||
label("<math>F</math>", (3.73,-7.39), NE * labelscalefactor); | label("<math>F</math>", (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); | ||
− | + | [/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>. |
Revision as of 19:30, 20 March 2024
Contents
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.