# "World's Hardest Easy Geometry Problem"

This question is a "corollary" (if you will) to the World's Hardest Easy Geometry Problem (external website). Formally, this is called Langley's Problem. The objective of that problem was to solve for angle $x^{\circ}$, with the given angles of $10^{\circ}, 70^{\circ}, 60^{\circ}, 20^{\circ}$. Someone presented a solution to that problem. Here's also a rather colorful and interactive solution to a problem like this, but with different angles.

Now, I wanted to generalize this problem, replacing the angles of $10^{\circ}, 70^{\circ}, 60^{\circ}, 20^{\circ}$ with angles of $W^{\circ}, X^{\circ}, Y^{\circ}, Z^{\circ}$, respectively (see below picture).

How can we derive an analytical expression of angle $x^{\circ}$, in terms of $W^{\circ}, X^{\circ}, Y^{\circ}, Z^{\circ}$?

• Coordinatize, and calculate approximately. Commented Apr 12, 2014 at 5:00
• That might work for the original problem with fixed degrees. But I'm not sure if placing the triangle on the coordinate grid would work for arbitrary angles $w^{\circ}, x^{\circ}, y^{\circ}, z^{\circ}$. Commented Apr 12, 2014 at 5:03
• You use $x^\circ$ for two distinct angles in your diagram... Commented Apr 12, 2014 at 5:04
• I got a good chuckle from the inclusion of "This figure is drawn to scale". Commented Apr 12, 2014 at 13:27
• I think the problem is more neatly stated without point C. We can just draw a quadrilateral and its diagonals, give four angles at the bottom, and say "find all the angles". Or put $|\overline{AB}|=1$ and say "solve this quadrilateral".
– h34
Commented Jun 26, 2015 at 15:42

Well, I'll give a guide to follow, not a final expression. I called you unknown angle as $\alpha_9$ (in order not to mess, because you have a big $X$ and a small $x$).
Since you know $x$ and $y$ then you know $\alpha_4$.
Since you know $x$, $w$, $z$ and $y$ then you know $\alpha_1$.
Since you know $x$ and $\alpha_5$ then you know $\alpha_3$.
Since you know $w$ and $\alpha_5$ then you know $\alpha_6$.
Then you end up with a system of $4$ equations: $$\begin{cases} \alpha_1+\alpha_2+\alpha_7=\pi \\ \alpha_4+\alpha_8+\alpha_9=\pi \\ \alpha_2+\alpha_3+\alpha_9=\pi\\ \alpha_6+\alpha_7+\alpha_8=\pi \end{cases}$$ with $4$ unknowns: $\alpha_2, \alpha_7, \alpha_8, \alpha_9$. And $\alpha_9$ is what you are looking for.

• It seems to me that using only relations between the angles results in an underdetermined system of linear equations. Commented Feb 11, 2015 at 16:56
• @ZhenLin - Start with line segment $\overline{AB}$ and draw lines upwards from A, making angles $x$ and $x+w$ with $\overline{AB}$, and from B, making angles $y$ and $y+z$ with it; all points are then determined.
– h34
Commented Jun 26, 2015 at 9:58
• That's not what I mean. If you actually use the geometry of the plane then there is no problem. Commented Jun 26, 2015 at 10:07
• Zhen Lin is correct, the 4x4 linear system here is underdetermined, and one needs to find another independent equation from the geometry to resolve the problem. To see this, take the first equation, add the second equation, and subtract the fourth equation. What you end out with is an equation of the form $\alpha_2 + \alpha_9 = [\text{known value}]$, which is of the same form as the third equation. Commented Jun 29, 2016 at 20:52
• But in this case the problem is that the matrix associated with the system of linear equations in four unknowns $\alpha_2, \alpha_7, \alpha_8$ and $\alpha_9$ given by \begin{pmatrix} 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \end{pmatrix} is not invertible. Commented Apr 20, 2020 at 7:38

I have found the equation!

I first tried to insert the latex equation directly here, but that was ugly so I rendered it separately and inserted an image instead:

• Now, how was this expression derived? :) Commented Apr 12, 2014 at 15:47
• My first attempt at solving it, I used only easy things like sum of all angles in triangle is $\pi$ and so on, but I couldn't solve it that way, so I decided to set the AB line to 1 and then write down what the other lines are in respect to that, when I had an expression for $a_8$, I wrote a computer program that expanded the expression to be in terms of the angles instead of the line lengths. The expression before the expansion was: $\arccos \left(\frac{DE^2 + IE^2 - DI^2}{2*DE*IE}\right)$ Commented Apr 12, 2014 at 20:25

There is in principle no problem in obtaining an expression for the top angle in terms of the bottom ones. Let the unlabelled intersection point in the picture be $I$. Let the bottom side be $1$. We can use the Sine Law to find $AI$ and $BI$. We can also find $AD$, and $BE$, and now we know $ID$ and $IE$, so we can solve for the mystery angle.

This does not result in a nice expression for the mystery angle, but it is an expression.

• This problem can be solved using geometry without also using trig. Commented Jun 29, 2016 at 13:51
• This is the correct answer Commented Jun 30, 2016 at 5:33

If one draws a perpendicular line from A and extends line DE to an intersection at, say X, the two resultant triangles allow the equal angles XEA and AEB to be determined at 40 degrees. Angle BED is then 100 degrees and 'x' is 30 degrees.

Thinking outside the triangle(s)?

• only for 10 degrees!
– user633401
Commented Jan 10, 2019 at 11:05

Refering to the Triangle Problem 1 diagram: If z>=w then x=80-w+arctan[(tan(z+10)-tan(w+10))/(tan(10)^2-tan(z+10)*tan(w+10))] degrees. There is no general solution for x in terms of z and w without using trigonometry since some solutions for x require an infinite series of digits to the right of the decimal point.

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– Community Bot
Commented Mar 14, 2022 at 9:55