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.

#Geometry #MathProblem #Math

Easy Math Editor

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Comments

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 - 7 years, 6 months ago

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

Sheikh Asif Imran Shouborno - 7 years, 6 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.

Jorge Tipe - 7 years, 6 months ago

For another good geometry problem go to https://brilliant.org/discussions/thread/geometry-problemagain-a-challenge/

Kishan k - 7 years, 6 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.

Ryan Soedjak - 7 years, 6 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!!

Ankit Chatterjee - 7 years, 6 months ago

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

Chandrachur Banerjee - 7 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$