Difference between revisions of "2024 USAMO Problems/Problem 5"
m |
(→Solution 1) |
||
| Line 6: | Line 6: | ||
== Solution 1 == | == Solution 1 == | ||
| + | define angle DBT as <math>/alpha</math>, the angle BEM as <math>/betta</math>. | ||
| + | Extend AD intersects BC at point T, then TC = TA, TE is perpendicular to AC | ||
| + | Thus, AB is the tangent of the circle BEM | ||
| + | |||
| + | Then the question is equivalent as the angle ABT is the auxillary angle of the angle BEM | ||
| + | as <math>/betta &= 180-B</math> | ||
==See Also== | ==See Also== | ||
{{USAMO newbox|year=2024|num-b=4|num-a=6}} | {{USAMO newbox|year=2024|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 08:14, 5 May 2024
- The following problem is from both the 2024 USAMO/5 and 2024 USAJMO/6, so both problems redirect to this page.
Contents
Problem
Point 
is selected inside acute triangle 
so that 
and 
. Point 
is chosen on ray 
so that 
. Let 
be the midpoint of 
. Show that line 
is tangent to the circumcircle of triangle 
.
Solution 1
define angle DBT as 
, the angle BEM as 
.
Extend AD intersects BC at point T, then TC = TA, TE is perpendicular to AC
Thus, AB is the tangent of the circle BEM
Then the question is equivalent as the angle ABT is the auxillary angle of the angle BEM as $/betta &= 180-B$ (Error compiling LaTeX. Unknown error_msg)
See Also
| 2024 USAMO (Problems • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 | ||
| All USAMO Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
