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.
