2014 USAJMO Problems/Problem 2
Contents
Problem
Let be a non-equilateral, acute triangle with , and let and denote the circumcenter and orthocenter of , respectively.
(a) Prove that line intersects both segments and .
(b) Line intersects segments and at and , respectively. Denote by and the respective areas of triangle and quadrilateral . Determine the range of possible values for .
Solution
Lemma: is the reflection of over the angle bisector of (henceforth 'the' reflection)
Proof: Let be the reflection of , and let be the reflection of .
Then reflection takes to .
is equilateral, and lies on the perpendicular bisector of
It's well known that lies strictly inside (since it's acute), meaning that from which it follows that . Similarly, . Since lies on two altitudes, is the orthocenter, as desired.
End Lemma
So is perpendicular to the angle bisector of , which is the same line as the angle bisector of , meaning that is equilateral.
Let its side length be , and let , where because lies strictly within , as must , the reflection of . Also, it's easy to show that if in a general triangle, it's equilateral, and we know is not equilateral. Hence H is not on the bisector of . Let intersect at .
Since and are 30-60-90 triangles,
Similarly,
The ratio is The denominator equals where can equal any value in except . Therefore, the denominator can equal any value in , and the ratio is any value in
Note: It's easy to show that for any point on except the midpoint, Points B and C can be validly defined to make an acute, non-equilateral triangle.
Solution 2
Let be the farthest point on the circumcircle of from line . Lemma: Line ||Line Proof: Set and , and on the unit circle. It is well known that and , so we have , so is real and thus the 2 lines are parallel.
WLOG let be in the first quadrant. Clearly by the above lemma must intersect line closer to than to . Intersect and at and and at . We clearly have , must intersect . We also have, letting the intersection of line and line be , and letting intersection of and be , . Since , and , also intersects . We have , so is equilateral. Letting , and letting the foot of the perpendicular from to be , we have , and since is an altitude of , we have . Letting the foot of the perpendicular from to be , we have by AA with ratio . Therefore, . Letting be the foot of the altitude from to , we have , since . Thus, since we have , so , so . We have , with , so can be anything in the interval . Therefore, the desired range is .
Solution by Shaddoll
See Also
2014 USAJMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |