The following is a geometry puzzle from a math school book. Even though it has been a long time since I finished school, I remember this puzzle quite well, and I don't have a nice solution to it.

So here is the puzzle:

alt text

The triangle $ABC$ is known to be isosceles, that is, $AC=BC$. The labelled angles are known to be $\alpha=\gamma=20°$, $\beta=30°$. The task is to find the angle labelled "?".

The only solution that I know of is to use the sine formula and cosine formula several times. From this one can obtain a numerical solution. Moreover this number can be algebraically shown to be correct (all sines and cosines are contained in the real subfield of the 36th cyclotomic field). So in this sense I solved the problem, but the solution is kind of a brute force attack (for example, some of the polynomials that show up in the computation have coefficients > 1000000). Since the puzzle originates from a book that deals only with elemetary geometry (and not even trigonometry if I remember correctly) there has to be a more elegant solution.

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    $\begingroup$ One geometric trick is add a few lines to create an equilateral triangle; then it's a straightforward geometric argument. $\endgroup$ Oct 16, 2010 at 13:24
  • $\begingroup$ @Robin: entering a return when commenting posts your comment. Just let it word wrap. I have made the same mistake. $\endgroup$ Oct 16, 2010 at 14:58
  • $\begingroup$ @Ross (and @Robin): Please ask the Stack Exchange people to fix this "feature": Pressing Enter in comment box unexpectedly submits form $\endgroup$ Oct 16, 2010 at 15:38
  • $\begingroup$ I wonder if there is a way to solve it for any alpha beta and gamma. $\endgroup$
    – Chao Xu
    Oct 16, 2010 at 16:07
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    $\begingroup$ I have changed the title. In future, I suggest you try to use a more appropriate title. Titles like "A geometry puzzle" are too generic. $\endgroup$
    – Aryabhata
    Oct 16, 2010 at 17:47

6 Answers 6


The solutions are not as trivial as one would expect from the statement. It's called Langley's problem of adventitious angles first posed in The Mathematical Gazette in 1922.

Check out An Intriguing Geometry Problem by Tom Rike.

  • $\begingroup$ the berkeley link is kind of confusing... but +1 for the name as a key to searching! $\endgroup$
    – Jason S
    Oct 16, 2010 at 14:35

I have written four distinct solutions to this problem: $$Solution Number 1:$$ enter image description here $$Solution Number 2:$$ enter image description here $$Solution Number 3:$$ enter image description here $$Solution Number 4:$$ enter image description here

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    $\begingroup$ I have a question in solution 1 : How do you know D' E AND C are collinear $\endgroup$
    – sidt36
    Feb 20, 2017 at 10:50
  • $\begingroup$ @sidt36, I have explained about it just under the first question. $\endgroup$
    – Seyed
    Feb 20, 2017 at 12:44
  • $\begingroup$ For solution 1 Will the same procedure work if make the angles and x,80-x and y,80-y $\endgroup$
    – sidt36
    Feb 20, 2017 at 15:04
  • $\begingroup$ Solution 1: I see that CN meets AF at right angles but don't see any bisector. How come $\angle D'FA=\angle FAD'$? And, as Blue commented to mina_world, please enter text as text (using MathJax for the maths); don't just display your text in your pictures. $\endgroup$
    – Rosie F
    Feb 6, 2019 at 12:21
  • $\begingroup$ Solution 2: If you want to define $O$ by $\angle CBO=20^\circ$, fine, but you'd better phrase it as $\angle CBO=\angle BAC$. Alternatively, say "let $O$ be on $CD$ where $OB=BC$". Then the solution is J.W. Mercer's. $\endgroup$
    – Rosie F
    Feb 6, 2019 at 12:25

Here is the solution I had for this (I had the writeup lying around in an old email I had sent regarding this, so, no tex, also, A and B are interchanged):

alt text

We can see that an $80-80-20$ triangle is nothing but a part of the triangulation of an $18$-sided regular polygon, whose $6$ triangles are shown in the the bottom part of the circle above.

Now consider an $80-80-20$ triangle ($ABC$ on top part of figure) and shoot a light ray from one of the base vertices ($B$ in the figure) at an angle of $50$ degrees to the base (or $30$ from one of the equal sides). (See the triangle in the top part of the circle and the red arrows)

We can show that the ray will reflect twice (first at $D$ then $E$) and strike at a $90$ degree angle the third time ($F$) i.e. after $5$ reflections, the light ray will return back to the vertex!

This reflection process can be pictured in another way, by reflecting the triangle each time instead of reflecting the ray (see the red arrows in the $6$ triangles in the bottom part).

Now the point of the third reflection $F$ (i.e. the $90$ degree incidence point) is exactly the midpoint of the side on which the ray is incident on. This can be seen by considering the bottom part:

Consider the right most $B$ and corresponding triangle $CBF$. This is a $90-60-30$ triangle. Thus $CF$ is half $CB$ which is half $CA$.

(Back to triangle at top) i.e. $F$ is midpoint of $AC$. Thus triangle $ACE$ is isosceles, thus angle $CAE = 20$.

Thus, we see that angle BDE must be the angle $x$ in the problem, which must be $180-(50+50) = 80$. (as $DE$ is $BD$ reflected on $AC$). That angle $y$ is $30$, follows…

Note: To get more context about what $x$ and $y$ are, this was the figure when the problem was shown to me:

alt text


This is the easiest solution. Don’t search for another.

Langley’s Adventitious Angles

  • Let $ D' $ lie on $ AC $ so that $ ED' \parallel BC $.
  • Let $ BD' $ meet $ CE $ at $ P $.
  • As $ \triangle BCP $ is equilateral, we have $ CB \equiv CP $.
  • As $ \triangle BCD $ is isosceles, we have $ CB \equiv CD $.
  • Hence, $ \triangle DCP $ is isosceles, which yields $ \angle CPD = 80^{\circ} $ and $ \angle DPD' = 40^{\circ} $.
  • As $ \angle DD'P = 40^{\circ} $ also, we find that $ \triangle D'DP $ is isosceles, so $ DD' \equiv DP $.
  • As $ \triangle D'EP $ is equilateral, we have $ ED' \equiv EP $.
  • Hence, by $ \mathsf{SSS} $, we obtain $ \triangle DED' \equiv \triangle DEP $.
  • Therefore, $ DE $ bisects $ \angle D'EP $, so $ \angle CED = \angle PED = 30^{\circ} $. $ \quad \blacksquare $

Any other solutions(advice) are welcome.

enter image description here

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    $\begingroup$ Math.SE has accumulated a number of duplicate $80^\circ$-$80^\circ$-$20^\circ$ questions over the years; it's refreshing to see someone post an answer. :) Please, though, enter the text of your solution as text (formatted using MathJax). Text in images is problematic; it isn't searchable, and it's useless to people who use screen readers. $\endgroup$
    – Blue
    Jan 23, 2019 at 4:39
  • $\begingroup$ @Blue thank you very much. $\endgroup$
    – mina_world
    Jan 23, 2019 at 4:45

This proof follows Seyed's first proof but with the point $D'$ constructed as the meet of $CE$ produced and the line through $B$ parallel to $AC$. It is clear that $\angle BD'C=30^\circ$. Recall that $\sin30^\circ=\frac12$, $\sin50^\circ=\cos40^\circ$, and $\sin80^\circ=2\sin40^\circ\cos40^\circ$. Then, applying the sine rule to triangles $BDC$ and $BD'C$ gives $$|BD'|=|BC|\frac{\sin50^\circ}{\sin30^\circ}=2|BC|\cos40^\circ=|BC|\frac{\sin80^\circ}{\sin40^\circ}=|BD|.$$ Hence triangles $BDE$ and $BD'E$ are congruent, and so $\angle BDE=\angle BD'E=30^\circ$.


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