1966 IMO Problems/Problem 2
Contents
[hide]Problem
Let , , and be the lengths of the sides of a triangle, and respectively, the angles opposite these sides. Prove that if
the triangle is isosceles.
Solution
We'll prove that the triangle is isosceles with . We'll prove that . Assume by way of contradiction WLOG that . First notice that as then and the identity our equation becomes: Using the identity and inserting this into the above equation we get: Now, since and the definitions of being part of the definition of a triangle, . Now, (as and the angles are positive), , and furthermore, . By all the above, Which contradicts our assumption, thus . By the symmetry of the condition, using the same arguments, . Hence .
Solution 2
First, we'll prove that both and are acute. At least one of them has to be acute because these are angles of a triangle. We can assume that is acute. We want to show that is acute as well. For a proof by contradiction, assume .
From the hypothesis, it follows that .
From it follows that . So,
because the numerator is (because for any real ), and the denominator is also (because , so ).
It follows that , so it can not be that .
Now, we will prove that implies .
Replace and (in fact, we don't care that is the radius of the circumscribed circle), and simplify by . We get
.
This becomes
We will show that the function is convex on the interval . Indeed, the first derivative is , and the second derivative is .
We have on since the numerator is (because for any real ), and the denominator is on the interval . It follows that is convex on the interval .
Using the convexity we have . In our case, we have
.
We can simplify by because it is positive (because both are acute!), and we get
. This is possible only when , i.e. .
[Solution by pf02, September 2024]
See Also
1966 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |