KVPY #23

Geometry Level 4

Consider a triangle $ABC$ whose side lengths are $AB = 4$ , $BC = 2$ and $CA = 3$ . If the bisector of $\angle C$ cuts $AB$ at $D$ and the circumcircle at $E$ , then find the value of $\left \lfloor \dfrac{CE}{DE} \right \rfloor$ .

Notation : $\lfloor \cdot \rfloor$ denotes the floor function .

The answer is 1.

2 solutions

Marta Reece
Apr 24, 2017

From law of cosines $\angle ACB=\arccos(\frac{2^2+3^2-4^2}{2\times2\times3})\approx104.5^\circ$

Because this is an obtuse angle, the center of the circumcircle, $O$ , is located outside the triangle $ABC$ and $DE>\frac{1}{2}CE$ .

This is all we need to know to determine that $\lfloor\frac{CE}{DE}\rfloor=1.$

It is certainly possible to calculate the value of the ratio, but it is not necessary.

that's a nice solution. It's a good idea.

Ahmad Saad - 4 years, 1 month ago

Nice simple solution. Voted up.
I did by long way, finding R and length CD, the angle bisector.

Niranjan Khanderia - 4 years, 1 month ago

Ahmad Saad
Apr 25, 2017

KVPY #23

The answer is 1.

2 solutions

1 pending report

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