Solve using basic Geometry!!(Challenge!)

In triangle ABC,AMABC,AM is the median such that MAB=105\angle MAB = 105^{\circ} and B=MAC\angle B = \angle MAC,find B\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
5 years, 11 months ago

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26 votes

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Great 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,CBACAM,CBA share the same angle C\angle C and B=MAC\angle B=\angle MAC, they are similar! Therefore their corresponding sides are proportional which gives: CMAC=ACCBAC2=CMAB=2CM2ABAM=ACCM=2\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 BAMBAM with BAM=105\angle BAM=105 and ABAM=2\frac {AB}{AM}=\sqrt {2}, find B\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 45459045-45-90 triangle? Let's try to construct one in this diagram! Construct a point NN on the same side as MM wrt ABAB such that ANB=90,BAN=45\angle ANB=90, \angle BAN=45. This means that NAM=10545=60\angle NAM=105-45=60. Since AN=AB2=AMAN=\frac {AB}{\sqrt {2}}=AM, we obtain that AMNAMN is an equilateral triangle. Furthermore, we have BN=AN=AM=NMBN=AN=AM=NM. Since BNM=90+60=150\angle BNM=90+60=150, therefore MBN=1801502=15\angle MBN=\frac {180-150}{2}=15 which means that B=4515=30\angle B=45-15=30! and that's our answer.

Xuming Liang - 5 years, 11 months ago

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Being too much exited, you made the answer factorial 30=30!30!. Ha ha ha. Good job.

Sheikh Asif Imran Shouborno - 5 years, 11 months ago

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

Motivation: Note that sinAMBsinABM=2\frac{\sin \angle AMB}{\sin \angle ABM }=\sqrt{2}, thus you need two angles whose sum is 7575^\circ and whose ratio of sines is 2\sqrt{2} ...yes! 45° and 30°.

Proof. Consider a triangle BAMB'A'M' such that ABM=30\angle A'B'M'=30^\circ and BMA=45\angle B'M'A'=45^\circ, note that ABAM=sin45sin30\frac{A'B'}{A'M'}=\frac{\sin 45^\circ}{\sin 30^\circ}, then ABAM=ABAM\frac{A'B'}{A'M'}=\frac{AB}{AM} and BAM=BAM=105\angle B'A'M'=\angle BAM=105^\circ, i.e. triangles BAMBAM and BAMB'A'M' are similar and we are done.

Jorge Tipe - 5 years, 11 months ago

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For another good geometry problem go to

Kishan k - 5 years, 11 months ago

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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.

Ryan Soedjak - 5 years, 11 months ago

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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!!

Ankit Chatterjee - 5 years, 11 months ago

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Great post dude. Need more of this kind to refresh minds

Chandrachur Banerjee - 5 years, 11 months ago

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