Problem:
Let △PQR be a right triangle with PQ=90,PR=120, and QR=150. Let C1​ be the inscribed circle. Construct ST, with S on PR and T on QR​, such that ST is perpendicular to PR and tangent to C1​. Construct UV with U on PQ​ and V on QR​ such that UV is perpendicular to PQ​ and tangent to C1​. Let C2​ be the inscribed circle of △RST and C3​ the inscribed circle of △QUV. The distance between the centers of C2​ and C3​ can be written as 10n​. What is n?
Solution:
Let r1​,r2​, and r3​ be the radii of circles C1​,C2​, and C3​, respectively. The inradius of any triangle is twice the area divided by the perimeter, so r1​=90⋅120/(90+120+150)=30. Because △RST is similar to △RPQ and RS=PR−2r1​=60, the similarity constant is 1/2. Thus r2​=15. Similarly, r3​=10. If d is the distance between the centers of C2​ and C3​, then
Let W,X, and Y be the points of tangency of circle C1​ to PQ​,PR, and RQ​, respectively. Note that PW=PX=r1​. Then RY=RX=120−r1​, and QY=QW=90−r1​, from which we obtain 120−r1​+90−r1​=150, and r1​=30. Assign coordinates so that P=(0,0),Q=(0,90), and R=(120,0). Now U=(0,60) and S=(60,0). Because triangles UQV and STR are similar to triangle PQR with similarity constants 1/3 and 1/2, respectively, conclude that r3​=10 and r2​=15. Thus the centers of circles C2​ and C3​ have coordinates (75,15) and (10,70), respectively. Use the distance formula to find that d2=652+552=7250.
