Problem:
Let ABC be a triangle and let ω be its incircle. Denote by D1​ and E1​ the points where ω is tangent to sides BC and AC, respectively. Denote by D2​ and E2​ the points on sides BC and AC, respectively, such that CD2​=BD1​ and CE2​=AE1​, and denote by P the point of intersection of segments AD2​ and BE2​. Circle ω intersects segment AD2​ at two points, the closer of which to the vertex A is denoted by Q. Prove that AQ=D2​P.
Solution:
The key observation is the following lemma.
Lemma Segement D1​Q is a diameter of circle ω.
Proof: Let I be the center of circle ω, i.e., I is the incenter of triangle ABC. Extand segement D1​I through I to intersect circle ω again at Q′, and extand segment AQ′ through Q′ to intersectt segment BC at D′. We show that D2​=D′, which in turn inplies that Q=Q′, that is, D1​Q is a diameter of ω.
Let ℓ be the line tangent to circle ω at Q′, and let ℓ intersect the seyments AB and AC at B′ and C′, respectively. Then ω is an excircle of triangle AB′C′′. Let H1​ denote the dialation with its center at A and ratio AD′/AQ′. Since ℓ⊥D1​Q′ and BC⊥D1​Q. ℓ⊥BC. Hence AB/AB′=AC/AC′=AD′/AQ′. Thus H1​(Q′)=D′.Hi​(B′)=B. and H1​(C′′)=C. It also follows that an excircle Ω of triangle ABC is tangent to the side BC at D′.
It is well known that
CD1​=21​(BC+CA−AB)(L)
We compute BD′. Let X and Y denote the points of tangency of circle Ω with rays AB and AC. respectively. Then by equal tangent.s. AX=AY⋅BD′=BX. and D′C=YC. Hence
AX=AY=21​(AX+AY)=21​(AB+BX+YC+CA)=21​(AB+BC+CA)
It follows that
BD′=BX=AX−AB=21​(BC=CA−AB)(2)
Combining (1) and (2) yields BD′=CD1​. Thus
BD2​=BD1​−D2​D1​=D2​C−D2​D1​=CD1​=BD′
that is, D′=D2​, as desired.
Now we prove our main result. Let M1​ and M2​ be the respective midpoints of segments BC and CA. Then M1​ is also the midpoint of segment D1​D2​, from which it follows that IM1​ is the midline of triangle D1​QD2​. Hence
QD2​=2IM1​(3)
and AD2​∥M1​I. Similarly, we can prove that BE2​∥M2​I.
Let G be the centroid of triangle ABC. Thus segments AM1​ and BM2​ intersect at G. Define transformation H2​ as the dialation with its center at G and ratio −1/2. Then H2​(A)=M1​ and H2​(B)=M2​. Under the dilation, parallel lines go to parallel lines and the intersection of two lines goes to the intersection of their images. Since AD2​∥M1​I and BE2​∥M2​I.H maps lines AD2​ and BE2​ to lines M1​I and M2​I. respectively. It also follows that H2​(I)=P and
APIM1​​=AGGM1​​=21​
or
AP=2IM1​(4)
Combining (3) and (4) yields
AQ=AP−QP=2IM1​−QP=QD2​−QP=PD2​
as desired.
Note: We used three different diagrams of triangle ABC. Each diagram was desgined to assist the reader in understanding a particular part of the proof. We used directed lengths of segements in our calculations to avoid possible complications caused by the different shapes of triangle ABC.
The problems on this page are the property of the MAA's American Mathematics Competitions
