enter image description here

Given the above problem, the solution seems trivial. Shouldn't it just be $$\dfrac{180^\circ-40^\circ-90^\circ }{2}=25^\circ ?$$

However because this is an olympiad problem, I think I might have gotten the answer wrong. How do you work this problem out? What am I doing wrong?

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    $\begingroup$ I don't think you can assume the angle adjacent to x is also x. $\endgroup$
    – angryavian
    Jan 27, 2023 at 5:57
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    $\begingroup$ you are wrong, the right answer is 30 degrees, hint: use trigonometric functions and the area of a triangle to figure it out. $\endgroup$
    – noname1014
    Jan 27, 2023 at 5:58
  • $\begingroup$ Why can't we assume this. Both angles are 10 degrees so shouldn't it bisect the side? Hence both angles should be x? $\endgroup$ Jan 27, 2023 at 6:15
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    $\begingroup$ No, it doesn't bisect the side. It would if it was perpendicular to it, but imagine the side falling closer and closer to the horizontal, and you can see how the upper "half" gets more and more large than then the bottom "half". $\endgroup$
    – JonathanZ
    Jan 27, 2023 at 6:39
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    $\begingroup$ @HaowenXie Have you heard of $\sin$ and $\cos$? Have you learned formulas such as $\sin(\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta$ and $2\sin\alpha\sin \beta=\cos(\alpha−\beta)−\cos(\alpha+\beta)$? $\endgroup$
    – Apass.Jack
    Jan 27, 2023 at 7:28

3 Answers 3


Considering the following picture, we use the fact that the sum of inner angles of a triangle is 180° to deduce the values 70° then 50° shown below. We then have the relations \begin{align*} b_1&=a_1\tan10=a_2\tan x\\ b_1+b_2&=a_1\tan20=a_2\tan50 \end{align*} from which we deduce $\displaystyle\tan x=\frac{\tan50\tan10}{\tan20}\cdot$ enter image description here It remains to simplify this expression. Let's use the formula $\displaystyle\tan\alpha\tan\beta =\frac{\sin\alpha\sin\beta}{\cos\alpha\cos\beta} =\frac{\cos(\alpha-\beta)-\cos(\alpha+\beta)}{\cos(\alpha-\beta)+\cos(\alpha+\beta)}$, with $(\alpha,\beta)=(50,10)$.

We get: $\displaystyle\tan50\tan10=\frac{\cos40-\cos60}{\cos40+\cos60} =\frac{2\cos40-1}{2\cos40+1}=\frac{4\cos^220-3}{4\cos^220-1}\cdot$

Since $\cos(3\alpha)=\cos\alpha(4\cos^2\alpha-3)$, we multiply numerator & denominator by $\cos20$ to obtain $\displaystyle\tan50\tan10=\frac{\cos60}{\cos20(4\cos^220-1)}$, and $\displaystyle\frac{\tan50\tan10}{\tan20} =\frac{\cos60}{\sin20(4\cos^220-1)}\cdot$

Finally, using the formula $2\sin\alpha\cos\beta=\sin(\alpha+\beta)+\sin(\alpha-\beta)$ we have \begin{eqnarray*} \sin20(4\cos^220-1)&=&2\cos20(2\sin20\cos20)-\sin20\\ &=&2\cos20\sin40-\sin20\\ &=&(\sin60+\sin20)-\sin20\\ &=&\sin60. \end{eqnarray*}

Conclusion: $\displaystyle\tan x =\frac{\tan50\tan10}{\tan20} =\frac{\cos60}{\sin60} =\tan30$, and $x=30$.

P.S.: the trigonometric computations are rather convoluted; they probably can be simplified but I didn't find out how…

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    $\begingroup$ Side note: From the formula $ \tan x \tan ( 60 ^ \circ - x) \tan (60^\circ + x) = \tan 3x$, substituting in $ x = 10 ^ \circ$ gives us $\tan 10^\circ \tan 50^\circ \tan 70^\circ = \tan 30^\circ$, which is what we want to deduce. $\endgroup$
    – Calvin Lin
    Jan 28, 2023 at 22:40

Please refer to the image below for a less complicated geometric solution.


As depicted in the picture, firstly, mirror the triangle $BDC$ across $BC$. Then mirror triangle $ABD'$ across $AB$.

Following the angles, you can see that $D''AC$ is a straight line. Furthermore, $\triangle BD'D''$ is equilateral and $\triangle BCD''$ is isosceles. Therefore $BD''=D'D''=CD''$, which implies that points $B$, $D'$ and $C$ all lie on a circle centred at $D''$. In other words, $D''$ is the circumcentre of $\triangle BD'C$.

Therefore $\angle BCD'=\frac12\angle BD''D'$.

Thus $x=30^{\circ}$.


I will also use this image https://i.sstatic.net/fAMYZ.png. Thanks.

The obvious thing is that $tg(20°)=(b1+b2)/a1$ and that $tg(50°)=(b1+b2)/a2$. And we see the common thing $b1+b2$. From this point, we can say that \begin{array}{l} tg(20°)*a1=tg(50°)*a2 \end{array} So, the connections of $10$ and $x$ are also the same (we have the same $b1$): \begin{array}{l} tg(20°)*a1=tg(50°)*a2\\ tg(10°)*a1=tg(x)*a2 \end{array} If we use $a2$ as a "connection point", we will obtain this: \begin{array}{l} (tg(20°)*a1)/tg(50) = (tg(10°)*a1)/tg(x)\\ tg(20°)/tg(50°) = tg(10)/tg(x)\\ tg(x) = (tg(50°) * tg(10°)) / tg(20°) \end{array} and x = 30


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