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 $\measuredangle AMC$?

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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:

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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 $\triangle ABC$

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

2 Answers 2


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.

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

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

  • $\begingroup$ I never saw the notation $X(a\mid b)$. What does it stand for!? Where can I learn more about!? $\endgroup$ Commented Jun 11, 2016 at 14:42
  • $\begingroup$ @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? $\endgroup$ Commented Jun 12, 2016 at 2:21
  • $\begingroup$ $M$ 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! $\endgroup$ Commented Jun 12, 2016 at 15:30

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