Problem:
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,CPQ, respectively. Given the fact that segment AP intersects ωA​,ωB​,ωC​ again at X,Y,Z respectively, prove that YX/XZ= BP/PC.
Solution:
First Solution: Assume that ωB​ and ωC​ intersect again at another point S (other than P). (The degenerate case of ωB​ and ωC​ being tangent at P can be dealt similarly.) Because BPSR and CPSQ are cyclic, we have ∠RSP=180∘−∠PBR and ∠PSQ=180∘−∠QCP. Hence, we obtain
∠QSR=360∘−∠RSP−∠PSQ=∠PBR+∠QCP=∠CBA+∠ACB=180∘−∠BAC;
from which it follows that ARSQ is cyclic; that is, ωA​,ωB​,ωC​ meet at S. (This is Miquel's theorem.)
Because BPSY is inscribed in ωB​,∠XYS=∠PYS=∠PBS. Because ARXS is inscribed in ωA​,∠SXY=∠SXA=∠SRA. Because BPSR is inscribed in ωB​,∠SRA= ∠SPB. Thus, we have ∠SXY=∠SRA=∠SPB. In triangles SYX and SBP, we have ∠XYS=∠PBS and ∠SXY=∠SPB. Therefore, triangles SYX and SBP are similar to each other, and, in particular,
BPYX​=SPSX​
Similar, we can show that triangles SXZ and SPC are similar to each other and that
SPSX​=PCXZ​
Combining the last two equations yields the desired result.
This problem and solution were suggested by Zuming Feng.
Second Solution: Assume that ωB​ and ωC​ intersect again at another point S (other than P ). (The degenerate case of ωB​ and ωC​ being tangent at P can be dealt with similarly.) Because BPSR and CPSQ are cyclic, we have ∠RSP=180∘−∠PBR and ∠PSQ=180∘−∠QCP. Hence, we obtain
∠QSR=360∘−∠RSP−∠PSQ=∠PBR+∠QCP=∠CBA+∠ACB=180∘−∠BAC;
from which it follows that ARSQ is cyclic; that is, ωA​,ωB​,ωC​ meet at S. (This is Miquel's theorem.)
Because BPSY is inscribed in ωB​,∠XYS=∠PYS=∠PBS. Because ARXS is inscribed in ωA​,∠SXY=∠SXA=∠SRA. Because BPSR is inscribed in ωB​,∠SRA= ∠SPB. Thus, we have ∠SXY=∠SRA=∠SPB. In triangles SYX and SBP, we have ∠XYS=∠PBS and ∠SXY=∠SPB. Therefore, triangles SYX and SBP are similar to each other, and, in particular,
BPYX​=SPSX​
Similar, we can show that triangles SXZ and SPC are similar to each other and that
SPSX​=PCXZ​
Combining the last two equations yields the desired result.
We consider the configuration shown in the above diagram. (We can adjust the proof below easily for other configurations. In particular, our proof is carried with directed angles modulo 180∘.)
Line RY intersects ωA​ again at TY​ (other than R ). Because BPYR is cyclic, ∠TY​YX= ∠TY​YP=∠RBP=∠ABP. Because ARXTY​ is cyclic, ∠XTY​Y=∠XAR=∠PAB. Hence triangles TY​YX and ABP are similar to each other. In particular,
∠YXTY​=∠BPA and BPYX​=PAXTY​​(1)
Likewise, if line QZ intersect ωA​ again at TZ​ (other than R ), we can show that triangles TZ​ZX and ACP are similar to each other and that
∠TZ​XZ=∠APC and PAXTZ​​=PCXZ​(2)
In the light of the second equations (on lengths proportions) in (1) and (2), it suffices to show that TZ​=TY​. On the other hand, the first equations (on angles) in (1) and (2) imply that X,TY​,TZ​ lie on a line. But this line can only intersect ωA​ twice with X being one of them. Hence we must have TY​=TZ​, completing our proof.
Comment: The result remains to be true if segment AP is replaced by line AP. The current statement is given to simplify the configuration issue. Also, a very common mistake in attempts following the second solution is assuming line RY and QZ meet at a point on ωA​.
This solution was suggested by Zuming Feng.
The problems on this page are the property of the MAA's American Mathematics Competitions
