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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. Xuming Liang · 3 years, 9 months ago

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@Xuming Liang 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. Jorge Tipe · 3 years, 9 months ago

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@Xuming Liang Being too much exited, you made the answer factorial 30=\(30!\). Ha ha ha. Good job. Sheikh Asif Imran Shouborno · 3 years, 9 months ago

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

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

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@Ryan Soedjak 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 · 3 years, 9 months ago

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Comment deleted Nov 20, 2013

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@Vikas Shukla 2 Vyom Chaturvedi · 3 years, 9 months ago

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@Vyom Chaturvedi Great reply if apples are short ask me i will gracefully provide Chandrachur Banerjee · 3 years, 9 months ago

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