# Secondary Selection 2023

Can someone provide the correct solution for the following problem? I tried to solve it with power of point theorem but keep seeing different answers from different people.

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I got 21\sqrt 3 and I didnt get any 1/4th power answer

Isn’t the answer \boxed{445} for a+b format?
I think the answer should be 4 \sqrt{21} . And then, 4\sqrt{\sqrt{21^{2}}} fits the given a\sqrt{\sqrt{b}} format.

Given,
In \triangle ABC, AD bisects \angle BAC into two equal halves, AB=18, AC=24, BP=12.

We get, \frac{BD}{DC} = \frac{AB}{AC}.

Now,
{{\frac{DC}{BD} =\frac{AC}{AB}}}

\implies \frac{DC}{BD}=\frac{24}{18}

\implies\frac{DC+BD}{BD}=\frac{24+18}{18}

\implies \frac{BC}{BD}=\frac{7}{3}

Now, by using power of a point for circumcircle of \triangle DEC, we get,

BP^2=BD\cdot BC

\implies BD\cdot BC=144

\implies \frac{3BC}{7} \cdot BC=144

\implies BC= \sqrt{144 \cdot \frac{7}{3}}

\implies BC=4 \sqrt{21}

\therefore \boxed{BC=4 \sqrt{21}}

(Please let me know if there is any mistake)

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