# Given equilateral $\triangle ABC$ and $M$ at distances $3$, $5$, $4$ from $A$, $B$, $C$, find $\angle AMC$.

Given that $$\triangle ABC$$ is equilateral triangle, and $$M$$ is a point inside $$\triangle ABC$$ such that $$AM=3\;\text{units} \qquad BM=5\;\text{units} \qquad CM=4\;\text{units}$$ what is the measure of $$\angle AMC$$ ?

I tried to use cosine rule since the sides are congruent, but it didn't help !

Any help is appreciated.

• Do you mean $$BM$$? Jul 2, 2019 at 15:30
• @ Dr. Sonnhard Graubner yes Jul 2, 2019 at 15:32
• The values does not fit the drawing!
– Moti
Jul 2, 2019 at 15:35
• @ Moti it is my fault , i fixed it Jul 2, 2019 at 15:39
• This one reminds me about a similar question that asks what is the AREA of the equilateral triangle.
– Moti
Jul 2, 2019 at 15:41

Reflect $$M$$ in $$BC$$ to obtain $$P$$. Reflect $$M$$ in $$CA$$ to obtain $$Q$$. Reflect $$M$$ in $$AB$$ to obtain $$R$$.

Then $$\triangle AQR$$, $$\triangle BRP$$ and $$\triangle CPQ$$ are three $$120^\circ$$-$$30^\circ$$-$$30^\circ$$ triangles. $$QR:PQ:RP=3:4:5$$. Hence $$\angle PQR=90^\circ$$. $$\angle AQR=\angle CQP=30^\circ$$. So, $$\angle AQC=150^\circ$$.

$$\angle AMC=\angle AQC=150^\circ$$.

• Thanks sir for this elegant solution, but how did you get $QR:PQ:RP=3:4:5$? Jul 3, 2019 at 3:00
• $AQR$, $BRP$, $CPQ$ are congruent triangles. So, $QR:PQ:RP=AQ:CP:BR$. We also have $AQ=AM$, $CP=CM$ and $BR=BM$. Jul 3, 2019 at 3:06

The hint.

Rotate $$\Delta MBC$$ on $$-60^{\circ}$$ around $$C$$.

I got $$\measuredangle AMC=90^{\circ}+60^{\circ}=150^{\circ}.$$

Hint:

This is a really long solution, but I think you'll get to the answer at the end.

Take the projections $$M_1$$, $$M_2$$, and $$M_3$$ of $$M$$ on $$AB$$, $$BC$$, and $$AC$$ respectively.

Now you have that $$AM^2= AM_1^2+ M_1M^2$$ and $$BM^2= M_1M^2+BM_1^2$$ and $$AM_1+BM_1=a$$

The same applies for all the 2 other triangles $$AMC$$ and $$BMC$$ (note that, $$AM_1+BM_1=AM_3+CM_3=CM_2+BM_2= a$$)

Now you get $$a$$ (the length of the side of the equileteral triangle) and then apply cosine law to get the angle.

(From the figure you can notice that $$5)