Difference between revisions of "2024 USAJMO Problems/Problem 1"
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==Solution 4 (Coord Bash)== | ==Solution 4 (Coord Bash)== | ||
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− | ~KevinChen_Yay | + | 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>. |
+ | |||
+ | Let's list what we know from the givens: | ||
+ | \begin{itemize} | ||
+ | \item 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}</cmath> and <cmath>c_1^2+c_2^2+d_1^2+d_2^2-2(c_1d_1+c_2d_2)=16.</cmath> | ||
+ | |||
+ | \item 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> | ||
+ | \end{itemize} | ||
+ | |||
+ | 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: | ||
+ | \begin{itemize} | ||
+ | \item <math>\overline{PS}</math>: | ||
+ | |||
+ | Notice that it has slope <math>-\frac{16a_1+12b_1-7c_1-21d_1}{16a_2+12b_2-7c_2-21d_2}</math> and point <math>(\frac{16a_1+12b_1+7c_1+21d_1}{28}, \frac{16a_2+12b_2+7c_2+21d_2}{28})</math>. 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}(x-\frac{16a_1+12b_1+7c_1+21d_1}{28}).</cmath> Since we claim that <math>(0,0)</math> in on the line, we substitute and 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 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> | ||
+ | |||
+ | |||
+ | "I will finish the proof when I 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 12:56, 28 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)
Let , , , .
Let's list what we know from the givens: \begin{itemize}
\item From the distance formula on and , we can simplify and get and
\item Since , , , and are weighted points on and , we can get along with
\end{itemize}
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: \begin{itemize}
\item : Notice that it has slope and point . The point-slope form equation would thus be Since we claim that in on the line, we substitute and get and simplifies to
"I will finish the proof when I 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.