HMMT November

Geometry Difficulty 4.9 AIME Find the answer

Let $A B C$ be a triangle with $A B=23, B C=24$ , and $C A=27$ . Let $D$ be the point on segment $A C$ such that the incircles of triangles $B A D$ and $B C D$ are tangent. Determine the ratio $C D / D A$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Let $X, Z, E$ be the points of tangency of the incircle of $A B D$ to $A B, B D, D A$ respectively. Let $Y, Z, F$ be the points of tangency of the incircle of $C B D$ to $C B, B D, D C$ respectively. We note that $C B+B D+D C=C Y+Y B+B Z+Z D+D F+F C=2(C Y)+2(B Y)+2(D F) 2(24)+2(D F)$ by equal tangents, and that similarly $A B+B D+D A=2(23)+2(D E)$ Since $D E=D Z=D F$ by equal tangents, we can subtract the equations above to get that $C B+C D-A B-A D=2(24)-24(23) \Rightarrow C D-D A=1$ Since we know that $C D+D A=27$ , we get that $C D=14, D A=13$ , so the desired ratio is $\frac{14}{13}$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.