Elegance is the key - Part III

Geometry Level 3

Consider a $\Delta ABC$ with incentre $I$ and circumcircle $\omega$ . Extend $\overline {AI}$ such that it meets $\omega$ at $M$ on the other side, as shown in the figure. Given that $\overline {MI} = 6$ units, find $\overline {MB} + \overline {MC}$ .

The answer is 12.

3 solutions

Adarsh Kumar
Aug 3, 2015

I have used simple properties of circles to find the angles shown in the picture below,just that the angles subtended by the same arc on different points on the circumference are equal and the exterior angle property. If you open the picture in a new tab,it would be much clearer.Sorry.

Moderator note:

Yup, this result can be arrived at through simple angle chasing.

Surya Prakash
Aug 3, 2015

Since $AI$ is the angular bisector.

$\angle MCB = \angle MAB = \angle MAC = \angle MBC$

So, $\Delta MBC$ is isosceles. So, $MC = MB$ .

$\angle MCI = \dfrac{1}{2} \angle BCA + \angle MCB = \dfrac{1}{2} \angle BCA + \angle MAB =$ $\dfrac{1}{2} \angle BCA + \dfrac{1}{2} \angle CAB = \angle MIC$ .

Since, $\angle MIC$ is the exterior angle of $\Delta AIC$ . So, $MI = MC$ .

So, $MC =MB = 6$ .

So, $MB + MC = \boxed{12}$ .

Moderator note:

Hm, I'm not too sure why you stated " $\angle MIC$ is the exterior angle of $\Delta AIC$ ". It seems to be that have shown $MIC$ is an isosceles triangle, hence $MI = MC$ .

This also illustrates why it is important to explain what each of your lines mean, and what you want to achieve.

Venkata Karthik Bandaru
Aug 3, 2015

[This solution has been deleted for being incorrect]

Moderator note:

I'm not sure how applying the converse theorem applies. Yes, the center of the circle will satisfy the stated angle relation, but that is not a unique point. Thus, how do we know that it must be the center of the circle? Hence, we are unable to conclude that $MI = MC = MB$ .

The converse of the theorem is true, and is quite widely used in proving concyclicity of points with respect to a given centre.

Given any 4 points in the plane such that the angle subtended by any two on the third point is twice that of the angle subtended on the fourth point, we can draw a circle passing through first, fourth, and second points, and the circle has centre as the third point.

Venkata Karthik Bandaru - 5 years, 10 months ago

Your comment, is very different from your solution.

It is true that if "Given 3 points A, B, C, and a fourth point D, then if $2 \angle ABC = \angle ADC, 2 \angle BCA = \angle BDA, 2 \angle CAB = \angle CDB$ , then D is the circumcenter of ABC."

What you are claiming in the solution is that "If $2 \angle ABC = \angle ADC$ , then D is the circumcenter of ABC. It should be obvious that this statement is not true, as there are 2 arcs which form the locus of point D, instead of just a single point.

Calvin Lin Staff - 5 years, 10 months ago

Thanks a lot sir, I now realise that I misinterpreted the converse. Thanks for being patient with me, and guiding me. I was extremely over confident. Cannot imagine a better Challenge Master than you for Brilliant !

Venkata Karthik Bandaru - 5 years, 10 months ago

Thanks a lot sir ! I was really stupid on that.

Venkata Karthik Bandaru - 5 years, 10 months ago

Elegance is the key - Part III

The answer is 12.

3 solutions

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