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.

This discussion board is a place to discuss our Daily Challenges and the math and science
related to those challenges. Explanations are more than just a solution — they should
explain the steps and thinking strategies that you used to obtain the solution. Comments
should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .

Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.

Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown

Appears as

*italics* or _italics_

italics

**bold** or __bold__

bold

- bulleted - list

bulleted

list

1. numbered 2. list

numbered

list

Note: you must add a full line of space before and after lists for them to show up correctly

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.

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.

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!!

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in`\(`

...`\)`

or`\[`

...`\]`

to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestGreat 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.

Log in to reply

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

Log in to reply

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.

Log in to reply

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

Log in to reply

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.

Log in to reply

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!!

Log in to reply

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

Log in to reply