I Am Proud Of This Tangent

Solution corresponding to how I found the problem:

For convenience let $BC=a, AC=b, AB=c$

According to the converse of alternate segment theorem , $\odot ADC$ is tangent to $CE$ (CE is a common tangent of two circles). By the same original theorem, $\angle ACE=\angle ADB=\angle AEB, \angle AEC=\angle ABE$ $\implies \triangle ABE\sim \triangle AEC$ , which means $AE$ bisects $\triangle BAC$ and has length $\sqrt {bc}$ . This solves the mystery of point $E$ .

We now take the tradition route to find $CD$ .

Let $AE\cap CB=F$ . Then $AF^2=bc(1-k^2)$ with $k=\frac {a}{b+c}$ . By power of a point , $\begin{aligned} BF\cdot DF &=AF\cdot EF \\ &=AF(AE-AF) \\ &=bc(\sqrt {1-k^2}-(1-k^2))\end{aligned}$

Since $BF=ck, CF=bk$ , therefore $DF=b(\frac {\sqrt {1-k^2}-(1-k^2)}{k}$ Hence $\boxed {CD=CF-DF=b(\frac {1-\sqrt {1-k^2}}{k})}$ .

It is given that $k=\frac {3}{5}, b=33$ ; plugging these above gives $CD=\boxed {11}$ .

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Remarks :

• If the problem were to ask for $BD$ , then the given conditions would not be enough.

• I suspect that a more clever solution exists, perhaps one that, through extra constructions, brilliantly uses the sum condition.

• I was really proud of this problem, so the title.