# A geometric puzzle. What is the angle?

This is a simply stated geometry puzzle. What is the angle p in this isosceles triangle?

Here's some information about the origin of the puzzle. Following any links therein may spoil the fun if you were planning to solve the puzzle yourself.

The puzzle was originally published in the Mathematical Gazette's puzzle/problem section in 1922; it was problem number 644, published in volume 11. It is due to Edward Mann Langley, a mathematics teacher and author, and has come to be known as Langley's Adventitious Angles. It's related to multiple intersections of diagonals in a regular 18-gon; see e.g. this puzzle here on Puzzling; the "adventitious" in the name is because it seems somehow arbitrary and random that the relevant three diagonals should meet at a point, or that all the angles in the quadrangle BCDE (including the ones formed by the diagonals) should be rational numbers of degrees.

• This is (I think -- I haven't checked it's exactly the same one as I'm thinking of) a famous -- and famously harder-than-it-looks-like-it-should-be -- puzzle. I wouldn't be at all surprised to find it's been asked here before, but if so finding the previous one might be difficult... – Gareth McCaughan Mar 31 at 14:56
• @GarethMcCaughan Could be. Please don't spoil the answer if you find it! – Anush Mar 31 at 14:57
• Well, if it's a duplicate then we need to mark it as such. But I haven't found another instance on the site yet... – Gareth McCaughan Mar 31 at 14:58
• Is ABC isosceles? – loopy walt Mar 31 at 14:59
• Now that a couple of answers have been posted, I have edited the question to say a bit about the source. – Gareth McCaughan Mar 31 at 19:59

Here is a geometric proof:

First, by tracing angles we can identify a couple of isosceles triangles, highlighted in colour:

M is the midpoint of AB
Angles: <ABD = <BAC = 20°
<ADX = <BDX = <DYX = 70°
[ <BCE = <BEC = 50° ] Not needed as far as I can tell.

Next, dropping perpendiculars from X to AD and DB construct the points P and Q. Observing that angle <ACX is 30°, we can see that 2 XP = 2 XQ = XC. Therefore, the configuration QXYCD is similar to MADCX. In particular, M is the midpoint of X and D, and triangle XED is isosceles.

The angle p is therefore

70° - 40° = 30°

Doing it the definitely hard way (and with some little cheating):

Let's introduce a coordinate system with $$BC=1$$ and origin in $$B$$. So the equation of $$BA$$ straight is $$y=\tan80^{\circ} x$$ (since $$\angle ABC=80^{\circ}$$). Similarly, $$CB$$ is $$y=-\tan80^{\circ} (x-1)$$ (because $$\angle ACB$$ is also $$80^{\circ}$$. For the same reasons we get $$CE$$: $$y=-\tan50^{\circ} (x-1)$$ and $$BD$$: $$y=\tan60^{\circ} x$$. Now we can find the coordinates of $$D$$ and $$E$$:
$$D=\left(\frac{\tan80^{\circ}}{\tan60^{\circ}+\tan80^{\circ}}, \frac{\tan60^{\circ}\tan80^{\circ}}{\tan60^{\circ}+\tan80^{\circ}}\right)$$
$$E=\left(\frac{\tan50^{\circ}}{\tan50^{\circ}+\tan80^{\circ}}, \frac{\tan80^{\circ}\tan50^{\circ}}{\tan50^{\circ}+\tan80^{\circ}}\right)$$
To find $$p$$, we can use the formula for angle between the vectors $$ED$$ and $$DB$$ (parentheses indicate scalar product): $$\cos p = \frac{(ED, DB)}{|ED|\times|DB|}=\frac{(x_E-x_D)x_D+(y_E-y_D)y_D}{\sqrt{(x_E-x_D)^2+(y_E-y_D)^2}\times\sqrt{x_D^2+y_D^2}}$$
Well, here must come the trig magic. Now, I've decided to cheat a bit and summon a sage... not, The Sage, plugging all the expressions and computing $$p$$ symbolically. Sage gave me that $$p = 30^{\circ}$$, which must be the answer.
Sage code (note that this gives 150 for $$p$$, probably due to sign error - it's clear that $$p$$ is an acute angle):


t50 = tan(5 * pi / 18)
t60 = tan(6 * pi / 18)
t80 = tan(8 * pi / 18)
xD = t80 / (t60 + t80)
yD = t60 * t80 / (t60 + t80)
xE = t50 / (t50 + t80)
yE = t50 * t80 / (t50 + t80)
c = (xD-xE)*xD + (yD-yE)*yD
g = sqrt((xE-xD)^2+(yE-yD)^2)
h = sqrt(xD^2+yD^2)
p = acos(c / (g+h)) * 180 / pi


• Can you include your sage code in the answer? – Anush Mar 31 at 16:55
• @Anush Thanks! I've included it. – trolley813 Mar 31 at 17:09