Solve the following problem....

Question

My problem is:

In a circle of radius $R$ is inscribed an equilateral triangle $ABC$. Through the point $C$ is drawn a line which intersects $AB$ in point $M$ and the circle, for the second time, in point $N$. Determine $CM\cdot CN$.

My idea is somehow to use sine theorem. So I wrote $CM$ as a function of $R$. But know I got stuck at finding $CN$ also as a function of $R$.

Thank you for help in advance!

The line drawn from C is any line that intersects AB and the circle ? — T_O, Mar 26 '14 at 10:05
Are you sure you don't need to give the expression of the product depending on some parameter, like the position of M on AB, or the angle $\hat{ACM}$ ? — T_O, Mar 26 '14 at 10:09
@JohnG. this is purely speculation but $M$ might be the mid-point of $AB$. also if this is homework, do you know what answer you're supposed to get? — Guy, Mar 26 '14 at 10:11
@Sabyasachi semi-unrelated but what is the length of AB given R ? — T_O, Mar 26 '14 at 10:12
@Sabyasachi So do I have to consider M as the mid-point AB? IN my problem is only written that the line intersects AB in M and that's all. — wonderingdev, Mar 26 '14 at 10:13
@T_O $\frac32 R = a\sqrt{\frac52}$ if I remember my geometry. — Guy, Mar 26 '14 at 10:15
@Sabyasachi No I don't have the result for this problem. It would be much more easier if I had. — wonderingdev, Mar 26 '14 at 10:20

ElThor · Accepted Answer · 2014-03-26T17:56:55.437

1

On the picture below you can find two similar triangles, one containing another.

Triangles $CBM$ and $CNB$ are similar because $\angle CBM = \angle CNB$ and $\angle BCM$ is the common angle.

After you write down proportions of the sides of those triangles you will see that $CM\cdot CN$ doesn't really depend on the position of point $M$ on $AB$. It also should give you a simple formula for $CM\cdot CN$ as a function of $BC$.

$CM\cdot CN = a^2$ where $a$ is the length of the side of $\triangle ABC$.
It follows from $\dfrac{CM}{CB} = \dfrac{CB}{CN}$.

It seems that only $R$ is given to you so you have to also express your answer as a function of $R$.

Note that $BC = a = R\sqrt{3}.$

image here

$CM\cdot CN = 3R^2$.

edited Mar 26 '14 at 17:56

answered Mar 26 '14 at 17:21

ElThor

912

Frankly , I do not get what are the similar triangles here. – wonderingdev Mar 26 '14 at 17:39
@JohnG. I tried to clarify my answer. Do you understand now? – ElThor Mar 26 '14 at 17:58
My another question is how do you know that angles N and B are the same (I mean those represented with green on you graph)? – wonderingdev Mar 26 '14 at 19:14
@JohnG. That can done in two steps. Firstly, $\angle B = \angle A$ because they're angles of an equilateral triangle. Secondly, $\angle A = \angle N$ because those are two inscribed angles intercepting the same arc. – ElThor Mar 26 '14 at 20:11

Solve the following problem....

1 Answers1