Interesting Circles

Geometry Level pending

In acute $\triangle ABC$ , let $D$ be the foot of the altitude from $A$ to $BC$ , and let $\overline{AD}$ intersect the circumcircle of $\triangle ABC$ again at $E$ . Let the circle with diameter $AE$ intersect lines $AB$ and $AC$ again at $N$ and $M$ , respectively. Given that $DB=3NB$ and $MA=5NA$ , then the value of $\frac{DC}{MC}$ can be written in simplest form as $\frac{a}{b}$ . What is the value of $a-b$ ?

6 14 -14 -2

1 solution

Aops Master
Feb 4, 2020

Notice that, since $E$ lies on the circumcircle of $\triangle ABC$ and $\angle EMA=\angle ENA=90^{\circ}$ , $M$ , $D$ , and $N$ are collinear (Simson Line). Then, by Menelaus's Theorem, we see that $\frac{NB}{NA}\cdot\frac{MA}{MC}\cdot\frac{DC}{DB}=1\rightarrow \frac{DC}{MC}\cdot\frac{NB}{DB}\cdot\frac{MA}{NA}=1$ , or $\frac{DC}{MC}\cdot\frac{5}{3}=1\rightarrow\frac{DC}{MC}=\frac{3}{5}$ , so our answer is $\boxed{-2}$ .

Interesting Circles

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