2013 Mock AIME I Problems/Problem 13
Problem
In acute , is the orthocenter, is the centroid, and is the midpoint of . It is obvious that , but does not always hold. If , , then the value of which produces the smallest value of such that can be expressed in the form , for squarefree. Compute .
Solution
Because and , we know that the height from to must be . Thus, because the perpendicular is the shortest segment from a line to a point not on the line, we know that . Thus, the minimum value of is , when . This is technically not possible, because is acute, but it may be helpful in getting a better sense of the problem. This scenario is shown below:
Because the orthocenter of is at , . Now, by the Pythagorean Theorem, . Because the centrod is of the way along the median from the vertex, we know that . Thus, we have , so the given inequality does not hold. Now, let us look at the example where is collinear with and :
By centroid properties, we know that . Let be the circumradius of . Then, by Pythagoras in , . Because and , we have the equation , which yields , so is between and . Because is acute, we know that is in the interior of . This fact, combined with the properties of the Euler Line, show that must be closer to than is, so , and the inequality is thereby satisfied.
As in the above diagram, let be the point on the circle such that is a diameter of the circle. Because is acute, and must be on the opposite sides of (so that the circumcenter lies inside the triangle). We want to be as close to as possible to minimize , but, from the first example we explored, we know that when , . We would reasonably expect the difference to vary continuously as we move towards , and this difference is positive in the second example and negative in the first example. Thus, by the Intermediate Value Theorem, there should be a point on the circumcircle between these two locations for such that this difference is zero, or . This point should be as close as we can get to while still satisfying the inequality.
Now, let and be the foot of the altitude from to . Further, let , so , as shown below:
By Pythagoras, we know that . Thus, by centroid properties, . Now, we desire to find another expression for . By using Pythagoras again, we see that and . Now, let . Also let and . By the Law of Cosines in , we have the following equation:
Because , . Plugging this into yields . Thus, our answer is .