Trisection is cool

Geometry Level 3

$ABC$ is a triangle such that $AC>AB$ , $D$ and $E$ are points on side $BC$ such that $BD=DC$ and $AE$ perpendicular to $BC$ . It is known that $AE$ and $AD$ trisect the angle $A$ and $BC=10\sqrt {3}$ . Find $AC$ .

The answer is 15.

2 solutions

Jayanta Mandi
Jun 15, 2014

The question is wrong.It should ask to find AC instead of AB to get the answer 15

Yeah you are right, it was a typing error.

mietantei conan - 6 years, 12 months ago

Kaan Dokmeci
Jun 15, 2014

First, we note $AD=AB$ as altitude $AE$ to $DB )\ bisects \( \angle{DAB}$ ; thus DAB is isosceles. Letting $\angle{ DAC=\theta }$ and applying the law of sines,

$\frac{\sin{\theta} }{5\sqrt{3}}=\frac{\sin{ C}{AD}$ to triangle ACD

$\frac{ \sin{3\theta}}{10\sqrt{3}}=\frac{\sin{C}{AB}$ to ACB.

But AD=AB so we have $2\sin{theta}=\sin{3\theta}$ , and inspection shows $\theta =30$ and BAC is right.

$AB$ is the shorter segment of the triangle $\triangle ABC$ . It is equal to $5\sqrt{3}$ , I'm quite sure the OP wanted us to find $AC$ instead (which is indeed equal to $15$ ). It could also be that he failed and got a wrong answer.

mathh mathh - 6 years, 12 months ago

I am very sorry I did an error in typing. I will edit the question right away.

mietantei conan - 6 years, 12 months ago

Trisection is cool

The answer is 15.

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