Problem:
Given a triangle ABC, let P and Q be points on segments AB and AC, respectively, such that AP=AQ. Let S and R be distinct points on segment BC such that S lies between B and R,∠BPS=∠PRS, and ∠CQR=∠QSR. Prove that P,Q,R,S are concyclic (in other words, these four points lie on a circle).
Solution:
We use the following lemma.
Lemma. Given a triangle ABC,X,Y,Z are points on BC,CA,AB respectively. Then three perpendicular lines of BC,CA,AB which go through X,Y,Z respectively are concurrent if and only if AY2+BZ2+CX2=AZ2+BX2+CY2.
Proof of Lemma. If the lines are concurrent, let P be the point on the three lines. From BX2−CX2=(PB2−PX2)−(PC2−PX2)=PB2−PC2 and so on, we obtain the desired result. Conversely, if AY2+BZ2+CX2=AZ2+BX2+CY2 holds, let Q be the intersection of perpendicular lines of BC,CA which go through X,Y respectively. Then as we have seen BX2−CX2=QB2−QC2 and CY2−AY2=QC2−QA2 holds. Summing up these equations, we have AZ2−BZ2=QA2−QB2. This implies that QZ and AB are perpendicular, as desired. End of the Proof
Let M be the midpoint of SR. We show that AP2+BM2+CQ2=AQ2+BP2+CM2. Since AP=AQ,CQ2=CR⋅CS,BP2=BS⋅BR, and BM2−CM2=(BM+CM)(BM−CM)=BC(BS−RC), we have (AP2+BM2+CQ2)−(AQ2+BP2+CM2)=BC(BS−RC)−BS⋅BR+CR⋅CS=BS⋅CR−CR⋅BC=0. Thus there exists a point O such that OP⊥BC,OQ⊥AC,OM⊥BC. Then O is the center of a circimcircle of PRS, since the circle is tangent to AB at P. Similarly, O is the center of a circumcircle of QRS, which implies that P,Q,R,S are on a circle.
Solution 2: By the given hypothesis, we have a circle Γ1 which passes through S and R, and touches AB at P. Similarly, we have a circle Γ2 which passes through S and R, and touches AC at Q. Suppose that the circles Γ1 and Γ2 are different from each other. Then the power of A onto Γ1 is AP2, and the power of A onto Γ2 is AQ2. This implies that A is on the radical axis of Γ1 and Γ2, namely the line BC, which is a contradiction. Hence, we have Γ1=Γ2, so that P,Q,R,S are concyclic, as desired.
Solution 3: We use the same notations as in the Solution 2. Suppose again that Γ1=Γ2. Let l be the perpendicular bisector of SR, and consider a circle γ passing through S and R whose center is moving on l. Suppose that initially the center of γ is on the half plane divided by BC in which A does not lie. Moving the center toward A,γ would touch AB and AC, not simultaneously by the hypothesis. Without loss of generality, suppose that γ touches AB at P first, and then touches AC at Q. Note that γ of these situations are Γ1 and Γ2 respectively.
We increases the radius of Γ1, keeping the circle tangent to AB. Then it will touch AC eventually. Let Γ1′ be the circle, which is tangent to AB and AC at P and Q respectively and meets BC at two points S′ and R′. Note that on BC, the points are ordered as B,S′,S,R,R′,C. We have ∠BPS=∠PRS and ∠BPS′=∠PR′S′, which imply ∠SPS′=∠RPR′. Similarly, we have ∠SQS′=∠RQR′. Without loss of generality, suppose that on the circle Γ1′, the points are ordered as S′,P,Q,R′. Let lines PS,PR,QS,QR meet Γ1′ again at T1,U1,T2,U2 respectively. Then the points on Γ1′ are
∠SQS′=∠RQR′, we have S′T2=U2R′. However, we have S′T2<S′T1=U1R′<U2R′,
which leads us to a contradiction. Hence, we have Γ1=Γ2, as desired.
Solution 4: Let Γ3 be the circle tangent to AB and AC at P and Q respectively. Inverse the plane around P. We denote by X′ the image of any point or any set X via the inversion. A′,P,B′ are collinear in this order, and the image of AC is a circle (AC)′ passing through A′ and P. Then Γ3′ is a line which is tangent to (AC)′ and parallel to A′P. Note that the tangency point is Q′. Γ1′ is a line parallel to A′P. Finally, B′,S′,R′ are on a circle passing through P, and S′,R′ are on Γ1′.
Suppose Γ1=Γ3. Then clearly we have Γ1′=Γ3′. Note that Q′ is on the perpendicular bisector l of A′P. Since PB′R′S′ is cyclic and PB′ and R′S′ are parallel, it is an isosceles trapezoid. Now we consider Γ2′. This circle should be tangent to Γ1′ at Q′, so the center of Γ2′ must lie on l. However, Since Γ2′ passes through R′ and S′, the center must lie on the perpendicular bisector of R′S′ which is the same as the one of PB′. Since A′ and B′ lie on the different ray centered on P, this is impossible. Therefore, we have Γ1=Γ3, on which P,Q,R,S lie.
Solution 5: In the case that AB=AC, suppose α=∠BPS>∠CQR=β. Let R′ be a point on BC such that BS=R′C. We then have that two triangles BPS and CQR′ are congruent. Hence, ∠CQR′=α>β=∠CQR, so that R lies between R′ and C. However, then we have β=∠QSC=∠PR′S>∠PRS=α, contradiction. Hence we have α=β, so the trapezoid PQRS is isosceles, as desired.
Now suppose AB=AC, and PQ and BC meet at X. Without loss of generality, suppose B>C so that B lies between X and C. Let AP=AQ=t,XB=x,BS=y,RC=z. To deduce x, we apply Menelaus' theorem to the triangle ABC and a line XPQ to obtain QCAQXBCXPABP=1. This yields x=b−cc−ta.
From the hypothesis, we have (c−t)2=y(a−z) and (b−t)2=z(a−y). From these results, we have (c−t)2−(b−t)2=(y−z)a, so that y−z=a(c−b)(b+c−2t). Hence, we obtain
On the other hand, since ∠APQ=2π−A, we have ∠PXB=2B−C. Applying the Sine theorem to the triangle XPB, we have sin2π−Ax=sinBXP⇔XP=xcos2AsinB. From Menelaus' theorem again, we have XPQXBAPBCQAC=1, or equivalently XQ=XPc−tcbb−t. Hence, we have
where R is the circumradius of the triangle ABC and S is the area of the triangle ABC. Since we have now that XP⋅XQ=XS⋅XR, the four points are concyclic, as desired.
Comment. It is a degenerated version of the following statement: if ABCDEF is a convex hexagon and ABCD,CDEF, and EFAB are cyclic quadrilaterals, then ABCDEF is a cyclic hexagon. This can be easily verified by the similar idea to the First and Second solution.
This problem and solution were suggested by Sungyoon Kim and Inseok Seo.
