69 of 100: No Trigonometry Required!

It's obvious that x = 45°. Mirror y vertically and try to figure out what the gap between y and z is. Looking at the grid you can see that AB = AC. Therefore in triangle ABC angle B = angle C. Also, noting that the slopes of AB and AC are negative reciprocals, you can see that angle A = 90°. Thus angle B = angle C = 45°. So x will perfectly fit into angle C. Conlusion: x + y + z = 180°.

Solution 1: Translation and Dilation

We can rearrange and resize the angles to get:

Solution 2: Addition through tangent identities

$\begin{aligned} \tan^{-1}(2)+\tan^{-1}(3) &= x\\ \tan(x) &= \tan (\tan^{-1}(2)+\tan^{-1}(3))\\ &= \dfrac{\tan (\tan^{-1}(2)) + \tan (\tan^{-1}(3))}{1- \tan (\tan^{-1}(2)) \tan (\tan^{-1}(3))}\\ &= \dfrac{2+3}{1-2 \times 3}\\ &= -1\\ x &= 135^{\circ}\\ \tan^{-1} (1) &= 45^{\circ}\\ \implies \tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3) &= \tan^{-1} (1) + x\\ &= 180^{\circ}\\ \end{aligned}$

Use complex numbers : $(1+i)(1+2i)(1+3i) = (-1+3i)(1+3i) = -1-9 = -10$ . In complex multiplication, the argument angles get added. The argument angle of $-10$ is \fbox{\$180^{\circ}\$} .

$\tan^{-1}3+\tan^{-1}2+\tan^{-1}1=\pi=180^{\circ}$

Proof: $\tan^{-1}3+\tan^{-1}2+\tan^{-1}1=\tan^{-1}\frac{3+2}{1-3\times 2}+\tan^{-1}1$ $=\tan^{-1}(-1)+\tan^{-1}1$ $=\tan^{-1}\frac{-1+1}{1+1}$ $=\tan^{-1}0$ $=\pi$ In the last step, we took the value of $\tan^{-1}0$ to be $\pi$ as the summation of those three anges must be greater than zero.

This is question # $69$ , and it is clear that the number $6$ is $9$ rotated $180^\circ$ . Mathematicians have a good sense of humour $\Rightarrow$ the answer is $\boxed{180^\circ}$ .

I rearranged the shaded angles and voila! A bit wonky, but nevertheless, a straight line which (as we all know) is 180 degrees.

I knew x = 45 degrees because it would be exactly half of 90 degrees. Then, even though we didn't have to use trigonometry, I used a inverse trig functions (and a calculator) to find tan-1 (1/2) to find part of what was missing in triangle y. Tan-1 (1/2) is about 27 degrees, so < y would be 180 - 90 - 27 or 63 degrees. Then I used the same process to find < z. Tan-1 (1/3) is about 18 degrees, so < z = 180 - 90 - 18 or 72 degrees. Lastly I added 45 + 72 + 63 to get 180 degrees.

x is the argument of the complex number 1+i

y is the argument of 1+2i

And z is the argument of 1+3i

When complex numbers are multipled together there arguments add together

(1+i) (1+2i)(1+3i) = (-1+3i) (1+3i) = -10

The argument of -10 is 180° so x+y+z=180° QED

Draw the square whose vertices are $(0,1)$ , $(1,3)$ , $(3,2)$ and $(2,0)$ (grid coordinates). The angle formed by one of the sides and the diagonal is $45^{\circ}=z-(90^{\circ}-y),$ whence $z+y=135^{\circ}$ and $x+y+z=180^{\circ}$ .

atan(1)+atan(2)+atan(3)=180

The line segments are rearranged and extended to form three sides of a triangle. In the process the three shaded angles have become the angles of a triangle. And we know that the sum of all the angles of a triangle is $180°$ .

Adjacent leg of each angle is 1; opposite of x = 1; opposite of y = 2; opposite of z = 3. Applying soh-cah-TOA, we need only find the arc-tangent of each quotient:
1 tan ^-1 = 45, 2 tan ^-1 = 63.43494882, 3 tan ^-1 = 71.56505118. Adding the three arc-tangents together, we get 180 degrees.

x = ${ 45 }^{ \circ }$

y = ${ tan }^{ -1 }(2)$

z = ${ tan }^{ -1 }(3)$

y + z = ${ tan }^{ -1 }(\frac{2 + 3}{1 - 2*3})$

= ${ tan }^{ -1 }(\frac{5}{-5})$

= ${ tan }^{ -1 }(-1)$

= ${ 135 }^{ \circ }$

x + y + z = 45 + 135

= ${ 180 }^{ \circ }$

$\angle x = 45^{\circ}$

angle between pink line and blue line + angle between pink line and green line = $45^{\circ}$

$\angle x + \angle y + \angle z = 3\angle x + 45^{\circ} = 180^{\circ}$

Look at Michel Van den Heuvel's solution. Good solution. I'd like to prove this somewhat rigouriously using no visuals but the same basic idea. (Trigonometric proof)

Without the picture, we would just have to prove that 1) triangle ABC has the same ratio of cathethi as the x-triangle i.e. that $\tan(\angle C)=1$ , and that

2) the triangle seen with those side lengths is actually right.

Name the given right triangles z-triangle, y-triangle and x-triangle respecticely.

1) This first result ist easily obtained, when you identify the construction of triangle ABC based on its hypotenuse and one cathesus. Inspired by the picture we carry on:

The pythagorean theorem provides that a right triangle with cathethis equaling the hypotenuse of the y triangle and hypotenuse equaling the cathehi of the y-triangle (that is ABC, right?) is of ratio 1.

These are $\sqrt{5}$ and $\sqrt(10)$ , And the resulting last cathethus is $\sqrt(10-5) = \sqrt(5)$

2) To prove that ABC is right, I'll show that the right triangle constructed on the hypotenuse of the x-triangle and with hypotenuse along the z-triangles hypotenuse necessarily has to have its hypotenuse equaling the hypotenuse of the z-triangle. So for this right triangle we have

$\cos(180-z-y) = \cos(90-v)$ for some $v \in (0°,90°)$ (because the triangle is right, right?)

Carrying on with the cosine relation $\cos(90-v) = \dfrac{ adjacent }{hypotenuse} = \dfrac{√5}{hyp}$

Now since the hypothenuse of the triangle is along the z-triangle, we have

$hyp = k \cdot √10$ for some real k.

Insert into the prior relation to get

$\cos(90-v) = \dfrac{√5}{k\cdot √10}$

$\Updownarrow$

$\cos(90-v) \cdot k√10= √5$

$\Updownarrow$

$\cos(90-v) \cdot k = \dfrac{1}{√2}$

And thus we get that $hyp = \sqrt{\frac{1}{2}}\sqrt{10}=\sqrt{\dfrac{10}{2}}=√5$

And we are done. We've shown that given ABC is a right triangle with those side lengths, it has the same ratio as the right x-triangle and thus the same angles, and also that a right triangle with the given hypotenuse of √5 must be exactly ABC.

X is 45, and something from geometry 10 years ago stuck because I knew y is 60 and z is 75. No complex math just something about the 3x3 grid.

Each angle is initially given as a vector with respect to the Cartesian plane. First find the unit vector for x, (rt2/2, rt2/2). Then convert the plane to these units and use the new plane to rotate to the vector (1, 2) in terms of these units. Normalize this to a unit vector and repeat the process with the following plane.

The problem statement "No trigonometry required" and the solution choices make it a giveaway. 180 degrees is the only possible solution option that could be solved for without trigonometry.

The required angle sum can be expressed as $x = arctan(1) + arctan(2) + arctan(3) \Rightarrow arctan(1) + arctan(2) = x - arctan(3).$ . If we take the tangent of both sides, then we obtain:

$tan[arctan(1) + arctan(2)] = tan[x - arctan(3)];$

or $\frac{tan(arctan(1)) + tan(arctan(2))}{1 - tan(arctan(1)) tan(arctan(2))} = \frac{tan(x) - tan(arctan(3))}{1 + tan(x) tan(arctan(3))};$

or $\frac{1 + 2}{1 - 2} = \frac{tan(x) - 3}{1 + 3tan(x)};$

or $-3 - 9tan(x) = tan(x) - 3;$

or $10tan(x) = 0 \Rightarrow tan(x) = 0 \Rightarrow x = arctan(0).$

Of the available choices, $x = \boxed{180}$ degrees works.

x is the argument of the complex number (1+i), y that of (1+2i) and z that of (1+3i).

x+y+z is the argument of (1+i)(1+2i)(1+3i) = -10

Therefore x+y+z = 180 °

tan y = 2, tan z = 3, tan(y + z) = (tan y + tan z)/(1 – tan y tan z), tan(y + z) = -1, y + z = 135 degrees. Therefore x + t + z = 45 + 135

= 180 degrees.

I know this isn't a solution from myself, but I thought some might enjoy this relevant video.

Numberphile Video - The Three Squares Geometry Problem

For what's it's worth, my solution to the problem was the one that used the $tan(A+B)$ formula.

Tangent (y+z) is equal to -1 (by tangent sum identity)

So (y+z) is equal to 135 degrees