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.
Solution 5
Extend AP to meet FE at point X. Note that there is symmetry of triangle across line AX. Therefore, if we can prove that the area of is twice the area of , then we will be done. For simplicity, let the side length of equilateral triangle be . \\
We can easily find the area of . Since the area formula for an equilateral triangle is , where is the side length, the area of is . By symmetry, the area of is exactly half the area of . Thus the area of .
Now we just need to find the area of . We have to find and for the area. Note that . It looks by eye that is equilateral. Let's try to prove it. \\
\textbf{\textit{Lemma 1: is equilateral.}} \\
\textbf{\textit{Proof of Lemma 1:}} By symmetry, across lines , we arrive to some conclusions: \begin{center}
. \end{center} Substituting the first equation into the second, we see that , which is necessary and sufficient to prove that is equilateral. \\ Since is equilateral, and so is , we find that . Because by symmetry across , we know that the similarity ratio is . Thus the side length of is , and . \\ Now we need to find . We can do . Well, that's easy! Since is a 30-60-90 triangle by symmetry across line , and , we know that . To find , we do pretty much the same thing: is a 30-60-90 triangle, and , so . We can now find : \begin{center} . \end{center} \\ Recalling why we were doing all this, we find the area of . It is . \\ \\ Let's summarize: \\ \begin{center} The area of is \\ The area of is , \\ which proves our initial claim. \end{center}
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 |