A Quidditch field is usually in the shape of an oval, a hundred and eighty feet wide and five hundred feet long. But today the Gryffindor house team is training on a smaller field in the shape of a triangle, with small towers in its corners $A,B,C$. The angles at $A$, $B$, $C$ are respectively $14^{\circ}$, $62^{\circ}$, $104^{\circ}$.

Harry is standing and resting in a point $H$ on the sideline $AC$ of the field, as he suddenly sees the Golden Snitch popping up at point $S$ on the sideline $AB$. Harry notes in a flash that $\angle HBC=50^{\circ}$ and that $\angle SCB=94^{\circ}$.

Question: How large is the angle $\angle CSH$?

(The answer to this question will be an integer. A good solution will clearly explain the reason why an integer number shows up here.)

  • 1
    $\begingroup$ Two superficially similar versions of this puzzle (with different angles) can be found here. I don't know how to solve those either, though. $\endgroup$ Commented Mar 22, 2016 at 16:27
  • $\begingroup$ @MichaelSeifert those two puzzles are very interesting and I think much easier than this puzzle because of the nicer angles and the ABC triangles being isosceles, which is not the case here. $\endgroup$
    – Ivo
    Commented Mar 22, 2016 at 17:37
  • 4
    $\begingroup$ To illustrate for other I made this image: i.sstatic.net/bkeY0.png. That drawing tool actually give me the answer, lol, but of course I won't just put it here without a step-by-step solution. $\endgroup$
    – Ivo
    Commented Mar 22, 2016 at 18:21
  • $\begingroup$ I have a feeling that we need to solve a system of equations involving $m\angle CSH$ and $m\angle HSC$; am I on the right track? $\endgroup$
    – DylanSp
    Commented Mar 22, 2016 at 20:14
  • 5
    $\begingroup$ @Oray I'm not sure voting down a question that you can't crack is considered very fair $\endgroup$ Commented Mar 23, 2016 at 22:36

6 Answers 6


Here is our situation. We are given the following setup:

enter image description here

I have omitted the "top" of the triangle (vertex $A$). We know the following angles:

$$ \begin{align} \angle ABC = \angle SBC &= u + v = 62^{\circ} \\ \angle ACB = \angle HCB &= s + t = 104^{\circ} \\ \angle HBC &= u = 50^{\circ} \\ \angle SCB &= t = 94^{\circ} \\ \end{align} $$

We wish to find:

$$ \angle HSC = x = {?}^{\circ} $$

We can solve for the unknown angles $w$, $y$, and $z$ by noticing two facts: the angles in a triangle add up to a constant ($\pi$) and the diagonals of the quadrilateral $HSBC$ create two pairs of triangles that share an angle. Therefore:

$$ \begin{align} t + u &= x + y \\ s + z &= v + w \\ \end{align} $$

To get a third equation, note that the total of the interior angles of any quadrilateral add up to $2\pi$:

$$ s + t + u + v + w + x + y + z = 2\pi $$

Solving this system of three equations gives us:

$$ \begin{align} w &= \pi - t - u - v \\ y &= t + u - x \\ z &= \pi - s - t - u \end{align} $$

Now, we can use the law of sines to write a second set of relations:

$$ \begin{align} \frac{\overline{HS}}{\sin s} &= \frac{\overline{CH}}{\sin x} & \frac{\overline{CH}}{\sin u} &= \frac{\overline{BC}}{\sin z} & \frac{\overline{BC}}{\sin w} &= \frac{\overline{SB}}{\sin t} & \frac{\overline{SB}}{\sin y} &= \frac{\overline{HS}}{\sin v} \end{align} $$

Multiplying them together, the lengths cancel:

$$ \sin s \sin u \sin w \sin y = \sin t \sin v \sin x \sin z $$

Adding in our substitutions from before (and remembering $\sin(\pi-x)=\sin x$):

$$ \sin s \sin u \sin (t+u+v) \sin (t+u-x) = \sin t \sin v \sin x \sin (s+t+u) $$

(This approach is adapted from the webpage Angular Angst that Fimpellizieri posted in this comment on the question.)

Now proving that the answer is $x=34^{\circ}$ (oops, did I say that out loud?) reduces to proving the following trigonometric identity:

$$ \sin 10^{\circ} \sin 50^{\circ} \sin 110^{\circ} \sin 156^{\circ} = \sin 12^{\circ} \sin 34^{\circ} \sin 94^{\circ} \sin 154^{\circ} $$

or equivalently:

$$ \begin{multline} \sin\left(\frac{\pi}{18}\right) \sin\left(\frac{5\pi}{18}\right) \sin\left(\frac{11\pi}{18}\right) \sin\left(\frac{13\pi}{15}\right) \\ = \sin\left(\frac{\pi}{15}\right) \sin\left(\frac{17\pi}{90}\right) \sin\left(\frac{47\pi}{90}\right) \sin\left(\frac{77\pi}{90}\right) \end{multline} $$

We can expand some of the fractions into some interesting sums:

$$ \begin{multline} \sin\left(\frac{\pi}{6}-\frac{\pi}{9}\right) \sin\left(\frac{\pi}{6}+\frac{\pi}{9}\right) \sin\left(\frac{\pi}{2}+\frac{\pi}{9}\right) \sin\left(\frac{13\pi}{15}\right) \\ = \sin\left(\frac{\pi}{15}\right) \sin\left(\frac{\pi}{6}+\frac{\pi}{45}\right) \sin\left(\frac{\pi}{2}+\frac{\pi}{45}\right) \sin\left(\frac{5\pi}{6}+\frac{\pi}{45}\right) \end{multline} $$

Remember that $\sin x=\sin(\pi-x)$; this allows us to put our identity into an even more intriguing form:

$$ \begin{multline} \sin\left(\frac{2\pi}{15}\right) \sin\left(\frac{\pi}{6}-\frac{\pi}{9}\right) \sin\left(\frac{\pi}{6}+\frac{\pi}{9}\right) \sin\left(\frac{\pi}{2}+\frac{\pi}{9}\right) \\ = \sin\left(\frac{\pi}{15}\right) \sin\left(\frac{\pi}{6}+\frac{\pi}{45}\right) \sin\left(\frac{\pi}{6}-\frac{\pi}{45}\right) \sin\left(\frac{\pi}{2}+\frac{\pi}{45}\right) \end{multline} $$

Let's see if we can't make a useful identity with the pattern that we see:

$$ \sin\left(\frac{\pi}{6}-\alpha\right) \sin\left(\frac{\pi}{6}+\alpha\right) \sin\left(\frac{\pi}{2}+\alpha\right) \\ = \left(\frac{1}{2}\cos\alpha - \frac{\sqrt{3}}{2}\sin\alpha \right)\left(\frac{1}{2}\cos\alpha + \frac{\sqrt{3}}{2}\sin\alpha \right)\cos\alpha \\ = \frac{1}{4}\left(\cos^2\alpha - 3\sin^2\alpha\right)\cos\alpha \\ = \frac{1}{4}\left(4\cos^3\alpha - 3\cos\alpha\right) \\ = \frac{\cos(3\alpha)}{4} $$

With this in mind, we can collapse our identity to:

$$ \sin\left(\frac{2\pi}{15}\right)\cos\left(\frac{\pi}{3}\right) = \sin\left(\frac{\pi}{15}\right)\cos\left(\frac{\pi}{15}\right) $$

This is beginning to look a lot more convenient. Using the double angle formula on $\sin(2\cdot\pi/15)$ and evaluating $\cos(\pi/3)$ we obtain:

$$ 2\sin\left(\frac{\pi}{15}\right)\cos\left(\frac{\pi}{15}\right)\frac{1}{2} = \sin\left(\frac{\pi}{15}\right)\cos\left(\frac{\pi}{15}\right) $$

Which is obviously true.

Since we know that the solution is unique, this proves that the solution is $34^{\circ}$. As to why, the only answer I can offer is "because it makes the beautiful cancellations above possible."

  • $\begingroup$ I don't have the patience to do this, but I would bet that if you use the formula $\sin X = \frac{e^{iX} - e^{-iX}}{2i}$ things will simplify. $\endgroup$ Commented Mar 25, 2016 at 2:51
  • $\begingroup$ @DaveBlackston Probably, I decided to go the trig substitution route instead. My guess is that the same key transformations would need to take place, but I'd have to use complex arithmetic rather than trig identities. Mathematica tells me that if I went that way I'd end up with $(-1)^{11/30}-(-1)^{19/30}$ on one side and $\frac{1}{2}\sqrt{7+\sqrt{5}-\sqrt{6(5+\sqrt{5})}}$ on the other... $\endgroup$ Commented Mar 25, 2016 at 3:09
  • 1
    $\begingroup$ I'm not quite sure what to make of this answer. It's incredibly beautiful: the cancellations are jaw-dropping. However, I'm not sure if it s rigorous. We know that $x$ exists, is unique and satisfies the product sines equality. I think what's missing is the implicit assumption that if some angle $\beta$ satisfies the equality, then $\beta = x$. In other words: $x=34°$ is a candidate (it satisfies the equality), but is it the only candidate? $\endgroup$ Commented Mar 25, 2016 at 10:29
  • 1
    $\begingroup$ I know I'm being overzealous; it's not so much about the answer being right or wrong as it is about the 'science' behind it. This is still very much incredible, and I'm amazed by the underlying structure your answer shows. $\endgroup$ Commented Mar 25, 2016 at 10:31
  • 1
    $\begingroup$ Eh, it's alright. The sine equation clearly has a single solution in $[0,\pi )$. Great, great answer! I guess this is another adventitious triangle. I might just have to verify that paper I linked earlier because something there is amiss. $\endgroup$ Commented Mar 25, 2016 at 10:43

Here is an incomplete solution that shows the answer, but not how to get it using elementary means. I also have no idea why the answer is an integral number of degrees.

The answer is 34 degrees.

As has been noted before, we can get almost all of the angles easily. Using this info and the laws of sines and cosines, we can go as follows. Label the interior point O and assume that the length BC is 1.

Calculate CH by using the law of sines on the triangle CBH.
$\sin(50)/CH = \sin(26)/1$

Similarly, calculate SB by using the law of sines on triangle CBS.
$\sin(94)/SB = \sin(24)/1$

Now, calculate SO and HO by using the law of sines on SOB and HOC.
$\sin(12)/SO = \sin(144)/SB$
$\sin(10)/HO = \sin(144)/CH$

Use the law of cosines to get SH.
$SH=\sqrt(SO^2 + HO^2 - 2*SO*HO*\cos(36))$

Finally, use the law of sines to get sin(X), and then X.
$\sin(X)/HO = \sin(36)/SH$

  • 1
    $\begingroup$ I have seen some answers pointing in this general direction, but unless a sufficientely good approximation is accepted as an integer, this type of answer fails to prove that the angle is exactly an integer degree (ie, an integer multiple of $\frac{\pi}{180}$). Trigonometry can be used to this effect, although the process is by no means easy. For instance, one might try to show that $\tan \left( \frac{15}{17} \theta \right) = \tan \left( \frac{\pi}{6}\right)$ -- that is, $\frac{15}{17} \theta$ is $30$ degrees -- using trigonometric identities. $\endgroup$ Commented Mar 24, 2016 at 2:07
  • $\begingroup$ FWIW, I agree with you. That's why I referred to my solution as incomplete. Some poking around seems to indicate that this problem (or at least similar problems) can be solved in a completely elementary fashion with no trigonometric identities required. I have no reason to believe that these suggestions are incorrect, but I don't really have a clue where to start with that approach. $\endgroup$ Commented Mar 24, 2016 at 2:13

I might be missing the point here. My answer is

36 degrees

I got this by

Remembering that all the angles in a triangle add up to 180 degrees, and that all the angles in a straight line add up to 180 degrees. (Also the fact that when 2 lines cross, opposite angles are the same, of course.) Then I drew a little picture and added and subtracted for a while.

I can't post the picture right now, but maybe tonight when I get home.

It is an integer because

All the given angles are integers. There is only addition and subtraction of integers involved in getting to the answer.

  • 1
    $\begingroup$ I know the answer and I can say that yours is not correct. Maybe you made a mistake with drawing. As far as I know you need to find X in this image: i.sstatic.net/bkeY0.png $\endgroup$
    – Ivo
    Commented Mar 22, 2016 at 18:24
  • $\begingroup$ That's a very nice drawing. Mine was just done in pen and ink, but looks about the same. Perhaps I made an error in my math. I will check again. $\endgroup$
    – Solocutor
    Commented Mar 22, 2016 at 18:37
  • $\begingroup$ @IvoBeckers: How did you make that drawing? $\endgroup$
    – dpwilson
    Commented Mar 22, 2016 at 18:38
  • $\begingroup$ I used geogebra.org . it's a very nice tool for making geometric images although it's not very straightforward how to make such angles and triangles i think so it requires some practice $\endgroup$
    – Ivo
    Commented Mar 22, 2016 at 18:57
  • $\begingroup$ My answer seems consistent with your drawing. I haven't found my error yet. (Though I can guess how it was made.) $\endgroup$
    – Solocutor
    Commented Mar 22, 2016 at 19:21

Made a mistake with my working! Following answer is wrong!

My answer is:


I drew out the diagram and with HSC labelled as x, I ended up with:

SHC = 38 + x

Then you can simply use (can't post the diagram at the moment)

180 = 10 + x + (38 + x)

Having found:

ASH = 156 - x because ASB = 180 and CSB = 180 - 62 - (104-10) = 24

to get the answer

  • 1
    $\begingroup$ How did you get SHC=38+x? That seems crucial to solving this puzzle. $\endgroup$
    – Trenin
    Commented Mar 23, 2016 at 13:46
  • $\begingroup$ This seems unlikely.  The drawing (which I independently recreated) shows an angle x that appears to be clearly ≤ 45°. $\endgroup$ Commented Mar 23, 2016 at 13:59
  • $\begingroup$ Ah, hold on - I've made a mistake with my working. My answer is completely wrong! $\endgroup$ Commented Mar 23, 2016 at 14:06

So using the image, I labelled the intersection in the middle $X$. Then I found the following angles:

In my notation, all angles are listed in clockwise order (i.e. $\angle ABC$ instead of $\angle CBA$) and longest sides are used (i.e. $\angle ABC$ instead of $\angle SBC$).

We are given: $$\angle CAB =14^{\circ}, \angle ABC=62^{\circ}, \angle BCA=104^{\circ}$$ $$\angle HBC=50^{\circ}, \angle BCS=94^{\circ}$$

From this, we can infer the following: $$\angle ABC = \angle HBC + \angle ABH \implies 62^{\circ} = 50^{\circ} + \angle ABH \implies \angle ABH = 12^{\circ}$$ $$\angle BCA = \angle BCS + \angle SCA \implies 104^{\circ} = 94^{\circ} + \angle SCA \implies \angle SCA = 10^{\circ}$$

Now we can infer one of the angles at $X$.

$$\angle CXB = 180^{\circ} - \angle BCS - \angle HBC = 180^{\circ} - 94^{\circ} - 50^{\circ} = 36^{\circ}$$

Thus, the opposite side of the intersection is the same and we can infer the other two angles at X.

$$\angle SXH = 36^{\circ}$$ $$\angle HXC = \angle BXS = 180^{\circ} - 36^{\circ} = 144^{\circ}$$

Now we can get some angles at $H$ and $S$. $$\angle CSB = 180^{\circ} - \angle ABC - \angle BCS = 180^{\circ} - 62^{\circ} - 94^{\circ} = 24^{\circ}$$ $$\angle CHB = 180^{\circ} - \angle HBC - \angle BCA = 180^{\circ} - 50^{\circ} - 104^{\circ} = 26^{\circ}$$

This is as far as I believe most people have gotten. I started using trig functions, but even if I find the answer, I will not be able to argue why it is an integer.


I believe the answer is

35 degrees


I won't bother with the angles since everyone seems to get it. Now I am not sure if this is mathematically valid (I doubt it) but I arbitrarily assigned BC a length of 100 assuming this triangle is possible and therefore all the side lengths will scale accordingly. With that in mind and using basic trig functions to figure out angles and sides of the smaller triangles, I got 34.5 and rounded it to 35.

  • $\begingroup$ My calculations (as well as Oray's and probably Ivo's) show an answer of 34 degrees, not 35. You probably have a rounding error somewhere. $\endgroup$ Commented Mar 23, 2016 at 21:30
  • $\begingroup$ Yeah probably but I'll just keep 35 up until I have a better calculator on hand. $\endgroup$
    – LampPost
    Commented Mar 23, 2016 at 21:39
  • $\begingroup$ By the way your method (assigning an arbitrary length) is totally valid, since the length has no effect on any of the angles. $\endgroup$ Commented Mar 23, 2016 at 21:47

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