Three Squares Problem

[bob012345]

TL;DR

Using only elementary geometry (not even trigonometry), prove that angle C in the figure equals the sum of angles A and B.

This problem is another from Martin Gardner’s book Mathematical Circus but originates from a fourth grade extra credit question in a Moscow school long ago. It makes me wonder how America won the Space Race! The problem statement is simply;

Using only elementary geometry (not even trigonometry), prove that angle C in the figure equals the sum of angles A and B.

There are apparently dozens of proofs of this so I suggest just showing your solution and others can come up with different solutions. I selected High School because it probably is at least that level for today’s readers. Have fun!

 

[Ibix]

Spoiler

The red triangle replicates the 3×1 triangle in the OP; the blue and green ones replicate the 2×1 triangle. These three triangles contribute one side each to the triangle $PQR$.

The angle $\angle PRQ$ is a right angle, since it is constructed from the green right angled triangle's $90-B$ angle and the blue triangle's $B$ angle. The sides $PR$ and $QR$ are manifestly equal lengths, so the triangle is right angled isoceles and hence angles $\angle QPR$ and $\angle PQR$ are 45°. From the angle $\angle QPR$, we can see that they are also equal to $A+B$, so $A+B=45°$. $C$ is also 45°, obvious from symmetry. Thus $C=A+B$.

BBC News once posted a challenging geometry problem from a Chinese maths exam taken aged 16, comparing it to the British exams taken at the same age.

A Chinese colleague observed that a Chinese child would have spent a lot of time learning to do similar puzzles and would therefore have solved it faster (using geometry) than I did (using vectors). But was the time spent learning to solve those time well spent? Did it give them maths skills, or just teach them tricks for solving geometry problems that will never come up in any practical situation?

 

[bob012345]

Here are some more solutions;

In each case, the triangle constructed in green is similar to the triangle highlighted in brown which has angle B and is added to angle A. The sum is equal to angle C.

 

[Gavran]

Using only elementary geometry (not even trigonometry), prove that angle C in the figure equals the sum of angles A and B.

Spoiler

$C+C=A+A+B+B\implies2C=2(A+B)\implies C=A+B$

 

[Lnewqban]

Just another approach:

 

[Steve4Physics]

The solution below is not mine – I saw it on YouTube, here. This is just my summary.

1. Add 3 squares and consider the triangle containing angle B:

2. Add 2 (reflected) copies of the triangle:

3. Note, below, that triangle PQR (red) is right-angled (at P) and isosceles, so has angles $90^{\circ} – 45^{\circ} – 45^{\circ}$.

4. From the original question we know angle C = $45^{\circ}$. So,inspecting the bottom left corner, we see that A + C +B = $90^{\circ}$ so A+B = C (= $45^{\circ}$).

 

[coquelicot]

 

[Gavran]

There are apparently dozens of proofs of this so I suggest just showing your solution and others can come up with different solutions. I selected High School because it probably is at least that level for today’s readers. Have fun!

Based on the previous posts, it can also be shown that A+B+C=D for the angles A, B, C, and D in the next picture.

Or A+B+C+D=E for the angles A, B, C, D, and E in the next picture.

 

[bob012345]

Here is another way but it’s just a variation of @coquelicot above in post #7 but the lengths are not obvious so they must be calculated.

Drop a perpendicular at point Q. Since point P is at the midpoint of the vertical line the distance PQ is $$\frac{1}{2\sqrt{2}}$$

Then the distance QM is $$\sqrt{2}-\frac{1}{2\sqrt{2}}= \frac{3}{2\sqrt{2}}$$
then the triangle formed from points PQM is similar to that formed with angle A. So angle PMQ is angle A but angle PMQ=A=C-B so A+B=C.

 

[Gavran]

TL;DR: Using only elementary geometry (not even trigonometry), prove that angle C in the figure equals the sum of angles A and B.

The equation $C=A+B$ is actually Euler's formula
$$ \arctan\frac12+\arctan\frac13=\frac\pi4 $$, and there are three more formulas of this kind.
Hermann's formula
$$ 2\arctan\frac12-\arctan\frac17=\frac\pi4 $$
, Hutton's or Vega's formula
$$ 2\arctan\frac13+\arctan\frac17=\frac\pi4 $$
, and Machin's formula
$$ 4\arctan\frac15-\arctan\frac1{239}=\frac\pi4 $$.
Hermann's formula and Hutton's formula can also be easily proved by using the approach presented in the above posts.

Hermann's formula

$\tan\measuredangle BAC=1/7$
$\tan\measuredangle DAE=1/2$
$\tan\measuredangle EAF=1/2$
$\measuredangle DAE+\measuredangle EAF-\measuredangle BAC=\measuredangle CAG$
$\tan\measuredangle CAG=1\implies\measuredangle CAG=\pi/4$

Hutton's formula

$\tan\measuredangle BAC=1/7$
$\tan\measuredangle DAE=1/3$
$\tan\measuredangle EAF=1/3$
$\measuredangle DAE+\measuredangle EAF+\measuredangle BAC=\measuredangle BAG$
$\tan\measuredangle BAG=1\implies\measuredangle BAG=\pi/4$

It would be hard to use this method in the case of Machin's formula because of $\arctan (1/239)$.

 

[bob012345]

Expanding the original problem to be an $nx1$ rectangle, it turns out the angle B to add to angle A (defined by the ratio $\large\frac{1}{n}$) to get $\pi/4$ is defined by the ratio $\large\frac{n-1}{n+1}$. Here is an example by construction for $n=4$.

First, construct a circle of radius $n+1$ centered on the origin and draw the line segment $y=\large\frac{x}{n}$ to intersect it.

Then construct a circle of radius $n-1$ centered on the intersection point.

Then construct a perpendicular line segment from that intersection point to the green circle.

Finally, draw the line segment from the origin to the point where that perpendicular line segment intersects the green circle. It will go through the point $(1,1)$ showing the sum of the angles is $\large\frac{\pi}{4}$. I believe this works for any integer $n$.

In the original problem angle A is defined by the ratio $\large\frac{1}{3}$ so $n=3$ so angle B is defined by the ratio $\large\frac{3-1}{3+1}=\large\frac{1}{2}$.