Difference between revisions of "Talk:2021 USAMO Problems/Problem 1"
m (typo) |
m (Latex formating fix) |
||
Line 9: | Line 9: | ||
Since <math>\angle A'AC = \frac{\pi}{2}</math>, <math>A'C</math> is a diameter of <math>X_3</math>, and <math>\angle A'OC = \frac{\pi}{2}</math>. Similarly, <math>\angle B_1OC = \angle C'OA = \angle B_2OA = \frac{\pi}{2}</math>, so O is on <math>B_2C', B_1A'</math>, and by opposite angles, <math>\angle A'OC' = \gamma = \angle A'CC'</math>. | Since <math>\angle A'AC = \frac{\pi}{2}</math>, <math>A'C</math> is a diameter of <math>X_3</math>, and <math>\angle A'OC = \frac{\pi}{2}</math>. Similarly, <math>\angle B_1OC = \angle C'OA = \angle B_2OA = \frac{\pi}{2}</math>, so O is on <math>B_2C', B_1A'</math>, and by opposite angles, <math>\angle A'OC' = \gamma = \angle A'CC'</math>. | ||
− | Finally, since <math>A'</math> is on <math>AA_2</math> and <math>C'</math> is on <math>CC_2</math>, that gives <math>\angle A_2AC' = \angle A_2CC_1</math> - meaning <math>C'</math> is on the line <math>AC_1</math>. But <math>C'</math> is also on the line <math>CC_1</math> - so <math>C'= | + | Finally, since <math>A'</math> is on <math>AA_2</math> and <math>C'</math> is on <math>CC_2</math>, that gives <math>\angle A_2AC' = \angle A_2CC_1</math> - meaning <math>C'</math> is on the line <math>AC_1</math>. But <math>C'</math> is also on the line <math>CC_1</math> - so <math>C'=C_1</math>. Similarly, <math>A'=A_2</math>. So the three diagonals <math>A_1C_2, B_1A_2, C_1B_2</math> intersect in <math>O</math>. |
[[File:USAMO_Q1.png|alt="USAMO Q1 graph"|800px|USAMO 2021 Q1 set-up]] | [[File:USAMO_Q1.png|alt="USAMO Q1 graph"|800px|USAMO 2021 Q1 set-up]] |
Latest revision as of 14:14, 15 September 2021
We are given the acute triangle , rectangles
such that
. Let's call
.
Construct circumcircles around the rectangles
respectively.
intersect at two points:
and a second point we will label
. Now
is a diameter of
, and
is a diameter of
, so
, and
, so
is on the diagonal
.
(angles standing on the same arc of the circle
), and similarly,
. Therefore,
.
Construct another circumcircle around the triangle
, which intersects
in
, and
in
. We will prove that
. Note that
is a cyclic quadrilateral in
, so since
,
is a diameter of
and
- so
is a rectangle.
Since ,
is a diameter of
, and
. Similarly,
, so O is on
, and by opposite angles,
.
Finally, since is on
and
is on
, that gives
- meaning
is on the line
. But
is also on the line
- so
. Similarly,
. So the three diagonals
intersect in
.