# What is the size of the angle $\angle AMC$? [duplicate]

Suppose we have a triangle $\triangle ABC$ where the size of two angles are given: $\measuredangle B=15^\circ$ and $\measuredangle C=30^\circ$. We draw the median $AM$, so now what is the size of angle $\measuredangle AMC$?

The answer is $45^\circ$... Ask your friends to think on it for a while...

Find different solutions... Different proofs...

This is my solution/proof:

Continue $CA$, draw the ray $\vec{CA}$. Then draw $BN$, in which it is perpendicular to $\vec{CA}$.

$\triangle BCN$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle. as we know about this triangle. $BN=\frac12 BC$. So $BN=BM$. This tells us $\triangle ABN$, is isosceles, so $\measuredangle BMN=\measuredangle BNM=\frac{180^\circ-\measuredangle MBN}2=\frac{120^\circ}{2}=60^\circ$.

• Now we know $\triangle BMN$ is equilateral.

Therefor $$MN=BN=BM\qquad(1)\\ \text{and }\measuredangle MBN=60^\circ\qquad{ }$$

By this, we get $\measuredangle CMN=180^\circ-\measuredangle BMN=180^\circ-60^\circ$, $$\text{therefor }\measuredangle CMN=120^\circ\ (2)$$

$\measuredangle ABN=\measuredangle CBN-\measuredangle{CBA}=60^\circ-15^\circ=45^\circ$. So, the other angle of the right triangle $\triangle ABN$, i.e. $\measuredangle BAN$, should be $180^\circ-90^\circ-45^\circ=45^\circ{}^\dagger$, therefor $\triangle ABN$ is isosceles.

So we would have $AN=BN$. By (1) we get $MN=AN$, so $\triangle AMN$ is isosceles. Therefor $$\measuredangle AMN=\measuredangle MAN=\frac{180^\circ-\measuredangle ANM}2=\frac{180^\circ-30^\circ}2=75^\circ$$

Now, by using (2) we can find the size of $\measuredangle AMC$, which is $$\angle AMC=\measuredangle CMN-\measuredangle AMN=120^\circ-75^\circ=45^\circ$$

${}^\dagger$let's note that one may also know that $\measuredangle BAN$ is equal to $45^\circ$, since it is an external angle of the triangle $\triangle ABC$

• How is median defined? There are a number of similar constructions such as angle bisector, and altitude. Commented Jun 10, 2016 at 2:38
• A median is any segment on a triangle with one point at the vertex and the other point bisecting the side opposite that vertex. Commented Jun 10, 2016 at 2:41
• Which contest?. Commented Jun 10, 2016 at 17:03

## 2 Answers

If we let $\alpha$ be the $\measuredangle BAM,$ then after strenuous calculation, we have that $$\alpha = \sin^{-1}\left(\sqrt{\frac{2 - \sqrt{3}}{2}}\right),$$ which we observe to be equal to $15$ degrees.

With this, we see that the desired angle is $180^{\circ} - 30^{\circ} - 15^{\circ} = \boxed{135^{\circ}}$.

And no, this is not worth a bounty.

• good job...but check it again... you got close... but not there yet... Commented Jun 10, 2016 at 2:53

Hint:
$A(0\mid 1)$
$B(-\cot{30°}\mid 0)$
$C(\cot{15°}\mid 0)$
$O(0\mid 0)$
What is the midpoint $M\,$ of $BC\,$?
What sort of triangle is $AOM\,$?

• I never saw the notation $X(a\mid b)$. What does it stand for!? Where can I learn more about!? Commented Jun 11, 2016 at 14:42
• @Omid Ghayour \\ My eyesight is not so acute anymore, so I use the vertical stroke to separate coördinates. Consider it as a harmless replacement for a comma. Have you noticed that in the title of this question you seek the angle AMC, whereas in the body of its text you seek the angle AMB? Commented Jun 12, 2016 at 2:21
• $M$ is the the midpoint of BC, i.e. BM=MC... AOM should be a right triangle, isn't so!? Thanks for remarking the mistake in the Title, I fixed it now! Commented Jun 12, 2016 at 15:30