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(This is just a question for fun. I saw a commercial logo today and I was inspired.

I have posted answers for this question and you may post alternative answers!)

Question

In the figure, $\triangle ABC$ is half of a square and $M$ is the midpoint of $BC$. Prove that $\alpha\neq\beta$.

Solution

$\triangle ABM$ and $\triangle AMC$ have the same area. They have a common side $AM$. Note that the area of either triangle is given by $S=\frac12(AB)(AM)\sin\alpha=\frac{1}{2}(AC)(AM)\sin\beta$. But $AB\neq AC$. So the equality holds only if $\alpha\neq\beta$.

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    $\begingroup$ This is an illustration that the tangent function is nonlinear. $\endgroup$ – user65203 Jul 21 '18 at 17:03
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    $\begingroup$ Fun question! Thanks for sharing :) $\endgroup$ – Sambo Jul 21 '18 at 17:09

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By contradiction: if the angles were equal, then $BM:MC=AB:AC$ (angle bisector theorem) and as a consequence $BM>MC$, which is false.

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A mostly visual proof:

enter image description here

Construct a second square with edge $BC$ as shown. Since $AC=BD$, $CM=BM$, and $\angle ACM$ and $\angle DBM$ are right angles, $\triangle ACM \cong \triangle DBM$.

Therefore $\angle CMA \cong \angle BMD$, so $A$, $M$, and $D$ are collinear. And $m\angle BDM = m\angle CAM = \beta$.

Now if $\alpha = \beta$, then $\triangle ABD$ is isosceles, with $AB = BD$. But this is plainly false since $AB = BD \sqrt{2}$. So $\alpha \neq \beta$.

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  • $\begingroup$ Beautiful! Anyone can understand immediately! $\endgroup$ – Sambo Aug 17 '18 at 20:45
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$\hspace{5cm}$enter image description here

If $\alpha=\beta$, then: $$1=\tan 45^\circ=\tan(\alpha+\beta)=\tan(2\beta)=\frac{2\tan\beta}{1-\tan^2\beta}=\frac{2\cdot \frac12}{1-\left(\frac12\right)^2}=\frac43,$$ hence a contradiction. So, $\alpha\ne \beta$.

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  • $\begingroup$ Nice! Seems that with your method we can generalize the question when $BC$ is divided into $n$ equal segments. $\endgroup$ – Mythomorphic Jul 22 '18 at 7:07
  • $\begingroup$ Nice observations both on commercial logo and on generalization! $\endgroup$ – farruhota Jul 22 '18 at 12:10
  • $\begingroup$ Haha, thank you! Not an advertisement, but the name of the company is Convoy. Take a look of it lol $\endgroup$ – Mythomorphic Jul 22 '18 at 12:15
  • $\begingroup$ I believe you should clarify that you're calculating tan β by dividing the edge sizes. Because for a moment it seemed that you replaced tan(22.5°) with 1/2 which is false :) $\endgroup$ – Pedro A Jul 22 '18 at 21:20
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Construct the circumcircle of $\triangle AMC$ and translate $\triangle BMA$ such that $BM$ coincides with $MC$. If the two angles were equal, $A'$ (the third vertex of the translated triangle) would lie on the circle, but it does not. Thus the two angles are not equal.

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  • $\begingroup$ Parcly.Very nice. $\endgroup$ – Peter Szilas Jul 21 '18 at 19:12
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Let $C'$ be the image of the orthogonal projection of $M$ onto $AB$. Suppose contrary that $\alpha=\beta$. Then, $MCA$ and $MC'A$ are congruent triangles. Thus, $MB=MC'$ and there is a contradiction here, which I will leave it as a mystery. In fact, you can use a similar argument to show that $\alpha<\beta$.

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Let line MD be constructed parallel to AC, with D on AB, so angle DMA = MAC. Now DM < DA, because DM < BD (Pythagorean Th.). But in any triangle, a greater side lies opposite a greater angle. Therefore angle DMA (that is, MAC) > BAM.

diagram

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From the diagram we have $$ \alpha + \beta = \pi/4 $$ If $\alpha = \beta$ this would give $\alpha = \beta = \pi / 8$.

Again from the diagram: Let the side of the square be $x$, then $$ \tan \beta = (x/2) / x = 1/2 $$ However $$ \tan(\pi/8) = \sqrt{2}-1 < 1/2 $$ So $\beta \ne \pi/8$ and $a \ne \beta$.

Appendix:

The addition theorem for the tangent yields $$ \tan(2x) = \frac{2\tan(x)}{1-\tan^2(x)} $$ So $$ \underbrace{\tan(2\cdot \pi/8)}_1 = \frac{2\tan(\pi/8)}{1-\tan^2(\pi/8)} $$ and we have the quadratic equation in $u = \tan(\pi/8)$: $$ \begin{align} 1-u^2 &= 2 u \iff \\ 1 &= u^2 + 2u \iff \\ 2 &= u^2 + 2u + 1 = (u+1)^2 \iff \\ u &= \pm\sqrt{2}-1 \end{align} $$ where we need the positive solution.

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Drop a perpendicular from $M$ to the line $AB,$ meeting $AB$ at $P.$ That is, $P$ is the point on line $AB$ closest to $M.$

Since $\angle ABM$ is not a right angle, $P$ is not $B,$ and therefore $MP < BM.$ Since $BM = CM,$ then $MP < CM.$

If $\alpha = \beta$ then the two right triangles $\triangle AMC$ and $\triangle AMP$ would be congruent, since all three angles and one side ($AM$) would be congruent. But since $MP < CM,$ the triangles are not congruent. Hence $\alpha \neq \beta.$

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By the law of sines, $$\sqrt{2}\frac{\sin(\alpha)}{MB}=\frac{1}{MA}=\frac{\sin(\beta)}{MC}.$$ Since $MC=MB$, we conclude that $\sqrt{2}\sin(\alpha)=\sin(\beta)$. Clearly neither $\alpha$ nor $\beta$ are integer multiples of $\pi$, so $\alpha\neq\beta$.

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First assume that $\alpha$ = $\beta$ and that both would be equal to $\frac{\pi}{8}^\circ$ (since $\alpha = \beta$ and $\alpha + \beta = \frac{\pi}{4} \to 2\beta = \frac{\pi}{4} \to \beta = \frac{\pi}{8}$).

If we call side $\overline{AC}$, n, then side $\overline{MC}$ is $\frac{n}{2}$. Next, we solve for $\overline{AM}$ in terms of n to show that $\alpha = \beta$ for all values n.

$\overline{AM}^2 = n^2 + (\frac{n}{2})^2 = n^2 + \frac{n^2}{4} = \frac{4n^2 + n^2}{4} = \frac{5n^2}{4}$

$\overline{AM} = \sqrt{\frac{5n^2}{4}} = \frac{\sqrt{5n^2}}{2} = \frac{n\sqrt{5}}{2}$ by the pythagorean theorem

Now that we have $\overline{AM}$, we can find $cos{\beta}$.

$\cos{\beta}=\frac{n}{\overline{AM}} = \frac{n}{\frac{n\sqrt{5}}{2}} = \frac{2n}{n\sqrt{5}} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$

$\cos{\beta} = \frac{2\sqrt{5}}{5}$

We have that $\cos{\beta} = \frac{2\sqrt{5}}{5}$ and we already established that $\beta = \frac{\pi}{8}$.

This would mean that $\cos{\frac{\pi}{8}} = \frac{2\sqrt{5}}{5}$. We can find $\cos{\frac{\pi}{8}} = \frac{\sqrt{2 + \sqrt{2}}}{2}$, which implies that $\frac{\sqrt{2 + \sqrt{2}}}{2} = \frac{2\sqrt{5}}{5}$.

If we square both sides, we get that $\frac{2+\sqrt{2}}{4} = \frac{20}{25} = \frac{4}{5}$. Then we can split the fraction, simplify, then subtract and simplify.

$\frac{2}{4} + \frac{\sqrt{2}}{4} = \frac{4}{5}$

$\frac{1}{2} + \frac{\sqrt{2}}{4} = \frac{4}{5}$

$\frac{\sqrt{2}}{4} = \frac{4}{5} - \frac{1}{2} $

$\frac{\sqrt{2}}{4} = \frac{3}{10}$

Next we can multiply both sides by 4 and conclude that...

$\sqrt{2} = \frac{12}{10} = \frac{6}{5}$

This is impossible since it has been proven that $\sqrt{2}$ is irrational and can't be written as a ratio of 2 integers. This would mean that $\sqrt{2} \neq \frac{6}{5}$ or $\frac{\sqrt{2 + \sqrt{2}}}{2} \neq \frac{2\sqrt{5}}{5}$ or $\cos{\frac{\pi}{8}}\neq \frac{2\sqrt{5}}{5}$

This means that $\beta \neq \frac{\pi}{8}$. Since $\alpha + \beta = \frac{\pi}{4}$, $\beta$ would have to be equal to $\frac{\pi}{8}$, but we proved that it cant be. This means that $\alpha \neq \beta$ since $\beta$ can no longer be multiplied by 2 to be $\frac{\pi}{4}$

This method works with tangent and sine as well (I believe it's a lot less work if I had used tangent instead of cosine, but this was how I first solved it)

I'm a high-school student and this is my first response, feedback is always appreciated!

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The figure $\triangle ABC$ is the half of a square therefore the sides $\overline{AB}$ and $\overline{BC}$ are of equal length. Now consider the triangle $\triangle ABM$. It is a rectangular triangle such as the original one.

Therefore the tangens of the angle $\alpha$ is given by

$$\tan(\alpha)~=~\frac{\overline{BM}}{\overline{AB}}$$

hence $M$ is the middle of the side $\overline{BC}$ this simplifies to

$$\tan(\alpha)~=~\frac{\overline{BM}}{\overline{AB}}~=~\frac{\frac{\overline{BC}}{2}}{\overline{AB}}~=~\frac{\frac{\overline{BC}}{2}}{\overline{BC}}~=~\frac{1}{2}\\ \alpha~=~\arctan\left(\frac{1}{2}\right)$$

Now go back to the triangle $\triangle ABC$. The tangens of the angle $\alpha + \beta$ is given by

$$\tan(\alpha + \beta)~=~\frac{\overline{BC}}{\overline{AB}}$$

which equals $1$. The tangens of the sum of two different angles is given by

$$\tan(x+y)~=~\frac{\tan(x)+tan(y)}{1-\tan(x)\tan(y)}$$

with $x=\alpha$,$y=\beta$ and $\tan(\alpha)=\frac{1}{2}$ you get

$$\begin{align} \tan(\alpha+\beta)~=~1~=~\frac{\frac{1}{2}+\tan(\beta)}{1-\frac{1}{2}\tan(\beta)}~\iff~1-\frac{1}{2}\tan(\beta)~&=~\frac{1}{2}+\tan(\beta)\\ \tan(\beta)&=\frac{1}{3}\\ \beta&=\arctan\left(\frac{1}{3}\right) \end{align}$$

Hence $\arctan\left(\frac{1}{2}\right)\neq\arctan\left(\frac{1}{3}\right)$ and so $\alpha\neq\beta$.

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Here's one intuitive way to analyze the problem:

If you're standing on point $A$ and looking at a car driving at constant speed on a straight line $BC$, you'll have to turn your head very fast when the car is closest to you (on point $C$) and slow the head rotation as the car drives away.

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Assume $\alpha$ and $\beta$ are both equal to $22.5^\circ$ (sum to forty-five).

The angle $\angle ABM$ must equal $45^\circ$ because it bisects the square.

The angle $\angle ACM$ occupies a square corner, so it must equal $90^\circ$.

$\alpha + \angle BMA$ must not sum to more than $135^\circ$ ($180^\circ$ $-$ $45^\circ$ from $\angle ABM$).

If alpha is equal to beta and both equal $22.5^\circ$, then BMA must equal $112.5$ degrees ($135^\circ - 22.5^\circ$).

BMA + AMC must equal less than $180$ degrees (one hemicircle); if $\angle BMA = 112.5^\circ$, then $\angle AMC$ must be $67.5^\circ$.

Given $\beta = \alpha = 22.5^\circ$, $\angle AMC + \angle ACM$ must equal $157.5^\circ$ ($180$ minus $\beta$); given the same assumption, $\angle AMC$ is equal to $67.5^\circ$ and sums with $\angle ACM$ to $147.5^\circ$.

Alpha must not be equal to beta.

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enter image description here

In the figure $$\alpha+b=m \tag{1}$$ Also, $$m+\beta=90^0\tag{2}$$ So from $(1)$ and $(2)$, $$\alpha+b=90^0-\beta\tag{3}$$ But, $$b=\alpha+\beta$$ So $(3)$ becomes $$2\alpha=90^0-2\beta$$ $$\implies \alpha=45^0-\beta$$

So we can see that $\alpha$ is $\beta$ less than $45^0$ so $\alpha$ and $\beta$ aren't equal.

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