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# Solve using basic Geometry!!(Challenge!)

In triangle $$ABC,AM$$ is the median such that $$\angle MAB = 105^{\circ}$$ and $$\angle B = \angle MAC$$,find $$\angle B$$? The problem looks simple but my challenge is to solve it using basic geometry(congruency,similarity etc) and not trigonometry.

Note by Kishan K
3 years, 9 months ago

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Great problem..man I haven't done a geometric proofs in such a long time, reminds me of the AoPS days:)

Here's a solution:

First of all since the triangles $$CAM,CBA$$ share the same angle $$\angle C$$ and $$\angle B=\angle MAC$$, they are similar! Therefore their corresponding sides are proportional which gives: $$\frac {CM}{AC}=\frac {AC}{CB} \Rightarrow AC^2=CM*AB=2CM^2 \Rightarrow \frac {AB}{AM}=\frac {AC}{CM}=\sqrt {2}$$.

So now the problem has been simplified to a more straight forward question: Given a triangle $$BAM$$ with $$\angle BAM=105$$ and $$\frac {AB}{AM}=\sqrt {2}$$, find $$\angle B$$ (Note that this triangle is "similarly" unique).

To proceed, let's get a bit "creative", what does the ratio of two sides equal to root 2 remind us of? A $$45-45-90$$ triangle? Let's try to construct one in this diagram! Construct a point $$N$$ on the same side as $$M$$ wrt $$AB$$ such that $$\angle ANB=90, \angle BAN=45$$. This means that $$\angle NAM=105-45=60$$. Since $$AN=\frac {AB}{\sqrt {2}}=AM$$, we obtain that $$AMN$$ is an equilateral triangle. Furthermore, we have $$BN=AN=AM=NM$$. Since $$\angle BNM=90+60=150$$, therefore $$\angle MBN=\frac {180-150}{2}=15$$ which means that $$\angle B=45-15=30$$! and that's our answer. · 3 years, 9 months ago

Nice solution! I also found that $$\frac{AB}{AM}=\sqrt{2}$$ using the same similarity, but finished with a "trick" solution.

Motivation: Note that $$\frac{\sin \angle AMB}{\sin \angle ABM }=\sqrt{2}$$, thus you need two angles whose sum is $$75^\circ$$ and whose ratio of sines is $$\sqrt{2}$$ ...yes! 45° and 30°.

Proof. Consider a triangle $$B'A'M'$$ such that $$\angle A'B'M'=30^\circ$$ and $$\angle B'M'A'=45^\circ$$, note that $$\frac{A'B'}{A'M'}=\frac{\sin 45^\circ}{\sin 30^\circ}$$, then $$\frac{A'B'}{A'M'}=\frac{AB}{AM}$$ and $$\angle B'A'M'=\angle BAM=105^\circ$$, i.e. triangles $$BAM$$ and $$B'A'M'$$ are similar and we are done. · 3 years, 9 months ago

Being too much exited, you made the answer factorial 30=$$30!$$. Ha ha ha. Good job. · 3 years, 9 months ago

For another good geometry problem go to https://brilliant.org/discussions/thread/geometry-problemagain-a-challenge/ · 3 years, 9 months ago

Great post dude. Need more of this kind to refresh minds · 3 years, 9 months ago

Hmm I got that no triangle exists... (by angle chasing and similar triangles)

EDIT: Oops I misread my paper, I'll post a solution soon. · 3 years, 9 months ago

I too got such a solution, and whats more, i got that the sum of angles of triangle ABC is greater than 210 degrees.all i did then was abandon this approach and look for another set of triangles where basic geo could be used!!and Lo it was there!! · 3 years, 9 months ago

Comment deleted Nov 20, 2013

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