2017 USAJMO Problems/Problem 3
Contents
Problem
() Let be an equilateral triangle and let be a point on its circumcircle. Let lines and intersect at ; let lines and intersect at ; and let lines and intersect at . Prove that the area of triangle is twice that of triangle .
Solution (No Trig/Bash)
Extend to hit at . Then note that Letting and , we have that Solving and simplifying using LoC on gives Similarly,
Now we find Note that Now let and . Then by an area/concurrence theorem, we have that or Thus we have that
Manipulating these gives Thus and we are done.
~cocohearts
Solution 1
WLOG, let . Let , and . After some angle chasing, we find that and . Therefore, ~ .
Lemma 1: If , then . This lemma results directly from the fact that ~ ; , or .
Lemma 2: . We see that , as desired.
Lemma 3: . We see that However, after some angle chasing and by the Law of Sines in , we have , or , which implies the result.
By the area lemma, we have and .
We see that . Thus, it suffices to show that , or . Rearranging, we find this to be equivalent to , which is Lemma 3, so the result has been proven.
Solution 2
We will use barycentric coordinates and vectors. Let be the position vector of a point The point in barycentric coordinates denotes the point For all points in the plane of we have It is clear that ; ; and
Define the point as The fact that lies on the circumcircle of gives us This, along with the condition inherent to barycentric coordinates, gives us
We can write the equations of the following lines:
We can then solve for the points :
The area of an arbitrary triangle is:
To calculate we wish to compute After a lot of computation, we obtain the following:
Evaluating the denominator,
Since and it follows that:
We thus conclude that:
From this, it follows that and we are done.
Solution 3
We'll use coordinates and shoelace. Let the origin be the midpoint of . Let , and , then . Using the facts and , we have , so , and .
The slope of is It is well-known that is self-polar, so is the polar of , i.e., is perpendicular to . Therefore, the slope of is . Since , we get the x-coordinate of , , i.e., . Using shoelace, So . Q.E.D
By Mathdummy.
Solution 4 Without the nasty computations
Note that . We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths.
Let and . Then, From Law of Cosine, .
From Ptolemy's theorem, , so .
Lemma 1: In Triangle ABC with side lengths and , the length of the angle bisector of is This can be easily proved with Stewart's and Law of Cosine.
Using Lemma 1, we have Plug in , we get: Then
By Mathdummy.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See also
2017 USAJMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |