EUCLIDEAN GEOMETRY

204 10. Complete Quadrilaterals

K∗
O

C
L∗

X L B
K T

AD

Figure 10.6B. Finding a hidden cyclic quadrilateral.

This is good, since we can apply our Lemma 9.18 now. Unfortunately, this does not
finish off the problem. We know that ∠ACB = 90◦ and CA is a bisector of ∠SCT , but we
actually want CT to bisect ∠ACB, or equivalently ∠SCT = 90◦.

The trick now is to consider radical axes. Since triangles XST and XAB are self-polar,
by Lemma 9.27 we find that O has the same power with respect to the circles with diameter

K∗
O

C
L∗

X L
K

S ADT B

Figure 10.6C. Completing the diagram for Example 10.16.

10.7. Problems 205

ST and AB. Hence the radical axis of the circles with diameter ST and AB contains the
point O. Moreover, the radical axis is perpendicular to the line through the centers, namely
AB. This implies it passes through C. Yet C lies on the circle with diameter AB. Hence it
lies on the circle with diameter ST as well, as desired.

Solution to Example 10.16. Let ωA and ωB be the circles through C centered at A and
B; extend rays AK and BL to hit ωB and ωA again at K∗, L∗. Evidently KLK∗L∗ is cyclic,
say with circumcircle ω. Moreover, by orthogonality we observe that AL, AL∗, BK, BK∗
are tangents to ω (in particular, KLK∗L∗ is harmonic).

This means that AB is the polar of X. Then D is the Miquel point of cyclic quadrilateral
KLK∗L∗, and it follows that T = KL∗ ∩ LK∗. This implies −1 = (K, K∗; L, L∗) =L
(S, T ; A, B) where S = KL ∩ K∗L∗. Hence it suffices to prove ∠SCT = 90◦.

As triangles XST and XAB are self-polar to ω, it follows that O has the same power
to the circles with diameter ST and AB. Hence the radical axis of these two circles is line
OC; this means C lies on the circle with diameter ST and we are done.

10.7 Problems

Problem 10.17 (NIMO 2014). Let ABC be an acute triangle with orthocenter H and let
M be the midpoint of BC. Denote by ωB the circle passing through B, H , and M, and
denote by ωC the circle passing through C, H , and M. Lines AB and AC meet ωB and
ωC again at P and Q, respectively. Rays P H and QH meet ωC and ωB again at R and S,
respectively. Show that BRS and CRS have the same area. Hints: 268 633 556

Problem 10.18 (USAMO 2013/1). In triangle ABC, points P , Q, R lie on sides BC,
CA, AB, respectively. Let ωA, ωB , ωC denote the circumcircles of triangles AQR, BRP ,
CP Q, respectively. Given the fact that segment AP intersects ωA, ωB , ωC again at X, Y ,
Z respectively, prove that Y X/XZ = BP /P C. Hints: 59 92 382 686

Problem 10.19 (Shortlist 1995/G8). Suppose that ABCD is a cyclic quadrilateral. Let
E = AC ∩ BD and F = AB ∩ CD. Prove that F lies on the line joining the orthocenters
of triangles EAD and EBC. Hints: 428 416 Sol: p.278

Problem 10.20 (USA TST 2007/1). Circles ω1 and ω2 meet at P and Q. Segments AC
and BD are chords of ω1 and ω2 respectively, such that segment AB and ray CD meet at
P . Ray BD and segment AC meet at X. Point Y lies on ω1 such that P Y BD. Point Z
lies on ω2 such that P Z AC. Prove that points Q, X, Y , Z are collinear. Hints: 277 615 525

Sol: p.279

Problem 10.21 (USAMO 2013/6). Let ABC be a triangle. Find all points P on segment
BC satisfying the following property: If X and Y are the intersections of line P A with the
common external tangent lines of the circumcircles of triangles P AB and P AC, then

P A 2 + P B · P C = 1.
XY AB · AC

Hints: 196 68 42 327

Problem 10.22 (USA TST 2007/5). Acute triangle ABC is inscribed in circle ω. The
tangent lines to ω at B and C meet at T . Point S lies on ray BC such that AS ⊥ AT . Points