Concyclic Points
8.4 THE POWER OF ELEMENTARY GEOMETRY 283Let's begin with a problem that explicitly asks us to find concyclic points. We will give
a leisurely treatment of two solutions, taking the time to reflect on strategies, tactics,
and tools.Example 8.4. 1 (USAMO 1990) An acute-angled triangle ABC is given in the plane.
The circle with diameter AB intersects altitude CC' and its extension at points M and
N, and the circle with diameter AC intersects altitude BB' and its extension at P and Q.
Prove that the points M, N, P, Q lie on a common circle.
Solution: We cannot overstate the importance of carefully drawing a diagram, and
keeping it as uncluttered as possible, at least intitially. The picture below contains noadditional objects, although we label the circles with diameters AC and AB by COl and
IDl, respectively.Our careful drafting--computer-assisted, in our case-immediately rewards us
with the observations that C' lies on COl and 8' lies on IDl. Are we just lucky? Ofcourse not! AB' B is a right triangle with right angle at 8' (the foot of altitude BB') with
hypotenuse AB, the diameter of IDl. By the inscribed right triangles fact (8.2.17), 8'
must lie on IDl. By the same reasoning, C' lies on COl. This was an easy geometric
observation, but it might have escaped our notice had our diagram been poorly drawn!
Now it is time to think strategically. We wish to prove that M, N, P, Q lie on a
circle. Let's list a few options, in increasing order of sophistication, for a penultimate
step.1. Look for a likely center of the alleged circle, and prove that the four points are
all equally distant from it.2. By the cyclic quadrilaterals lemma (8.2.18), if we can show that angles QMP
and QNP are supplementary, we're done.3. Using the same lemma, another approach is to show that LQPM = LQNM.
4. Use the converse of the Power of a Point theorem (Problem 8. 3 .29) to deduce
that M, N, P, Q are concyclic. We will need to label the intersection of PQ and