Problem:
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 PA with the common external tangent lines of the circumcircles of triangles PAB and PAC, then
(XYPA​)2+AB⋅ACPB⋅PC​=1
Solution:
We consider the left-hand side configuration shown below. Let OB​ and ωB​(OC​ and ωC​) denote the circumcenter and circumcircle of triangle ABP(ACP) respectively. Line ST, with S on ωB​ and T on ωC​, is one of the common tangent lines of the two circumcircles. Point X lies on segment ST. Point Y lies on the other common tangent line.
We will start with the following simple and well known geometry facts.
Let M be the intersection of segments XY and OB​OC​. By symmetry, M is the midpoint of both segments AP and XY, and line OB​OC​ is the perpendicular bisector of segments XY and AP. By the power-of-a-point theorem,
XS2=XAâ‹…XP=XT2 and X is the midpoint of segment ST(3)
Triangles ABC and AOB​OC​ are similar to each other, which is the so called Salmon theorem. Indeed, ∠ABC=∠MOB​A=∠OC​OB​A, because each angle is equal to half of the angular size of arc \overparen{A P} of ωB​. Likewise, ∠OB​OC​A=∠C. In particular, we have
AOB​AB​=OB​OC​BC​=OC​ACA​(4)
Set AB=c,BC=a, and CA=b. We will establish the following key fact in two approaches.
1−(XYPA​)2=(AB+AC)2BC2​=(b+c)2a2​(5)
With this fact, the given condition in the problem becomes
AB⋅ACPB⋅PC​=(b+c)2a2​ or PB⋅PC=(b+c)2a2bc​(6)
There are precisely two points P1​ and P2​ (on segment BC ) satisfying (6): AP1​ is the bisector of ∠BAC and P2​ is the reflection of P1​ across the midpoint of segment BC. Indeed, by the angle-bisector theorem, P2​C=P1​B=b+cac​ and P2​B=P1​C=b+cab​, from which (6) follows.
In order to settle the question, it remains to show that we can't have more than two points satisfying (6). We just write (6) as
(b+c)2a2bc​=PB⋅PC=PB⋅(a−PB)
This a quadratic equation in PB, which can have at most two solutions.
. Rays OB​X and OC​T meet in W. Because of (3) and OB​S∥OC​T, triangles OB​SX and WTX are congruent to each other. Hence OB​X=XW and triangles OB​XOC​ and WXOC​ have the same area. Note that XM and XT are altitudes in triangles OB​XOC​ and WXOC​ respectively. Hence
4XY⋅OB​OC​​=2XM⋅OB​OC​​=2XT⋅OC​W​=4ST⋅(OC​T+TW)​=4ST⋅(OC​T+OB​S)​
By (4), we can write the above equation as
STXY​=OB​OC​OC​T+OB​S​=OB​OC​OC​A+OB​A​=BCAB+AC​ or ST2XY2​=a2(b+c)2​(7)
Note that OB​STOC​ is a right trapezoid. Let U be the foot of the perpendicular from OC​ on OB​S. We have
ST2=UOC2​=OB​OC2​−OS​U2=OB​OC2​−(OB​S−OC​T)2=OB​OC2​−(OB​A−OC​A)2
By (4), we can write the above equation as
ST2=BC2OB​OC2​​(BC2−(BA−CA)2)=BC2OB​OC2​​(a2−(b−c)2)=BC2OB​OC2​​(a+b−c)(a−b+c)(8)
Multiplying (7) and (8) together gives
XY2=BC2OB​OC2​​⋅a2(a+b−c)(a−b+c)(b+c)2​(9)
Let ha​ denote length of the altitude from A to side BC in triangle ABC. Then ha​ and AM are corresponding parts in similar triangles ABC and AOB​OC​, and so
BC2OB​OC2​​=ha2​AM2​=4ha2​AM2​(10)
Multiplying (9) and (10) together gives
XY2=4ha2​AP2​⋅a2(a+b−c)(a−b+c)(b+c)2​
By Heron's formula, we have
XY2AP2​=(a+b−c)(a−b+c)(b+c)24ha2​a2​=(b+c)2(a+b+c)(b+c−a)​=(b+c)2(b+c)2−a2​=1−(b+c)2a2​
from which (5) follows.
Solution 2. By the power-of-a-point theorem, we have XAâ‹…XP=XS2. Therefore,
1−(XYPA​)2=XY2XY2−PA2​=XY2(XY+PA)(XY−PA)​=XY24XA⋅XP​=XY24XS2​XY2ST2​(11)
Let S1​ and T1​ be the feet of the perpendiculars from S and T to line OB​OC​. It is easy to see that right triangles OB​SS1​,OC​TT1​,OS​OC​U are similar to each other. Note also that XM is the midline of right trapezoid S1​STT1​ (because of (3)). Therefore, we have
OB​OC​ST​=OB​OC​UOC​​=OB​SS1​S​=OC​TT1​T​=OB​S+OC​TS1​S+T1​T​=OB​S+OC​T2XM​=OB​S+OC​TXY​
or, by (4),
XYST​=OB​S+OC​TOB​OC​​=OB​A+OC​AOB​OC​​=BA+CABC​=b+ca​(12)
It is clear that (5) follows from (11) and (12).
This problem and were suggested by Titu Andreescu and Cosmin Pohoata. Solution 2 was suggested by Zuming Feng.
The problems on this page are the property of the MAA's American Mathematics Competitions
