(Seemingly) simple trigonometry problem

Solution 1:

The trick here is to shift the perpendicular between the two diagonal lines upwards so that two similar triangles are obtained. Then, by applying the Pythagorean theorem, we obtain $$\frac ax=\frac b{\sqrt{b^2+(c-x)^2}}$$ which may be manipulated to obtain a quadratic polynomial for $x$: $$a\sqrt{b^2+(c-x)^2}=bx$$ $$a^2(b^2+(c-x)^2)=b^2x^2$$ $$b^2+c^2-2cx+x^2=\frac{b^2}{a^2}x^2$$ $$\left(1-\frac{b^2}{a^2}\right)x^2-2cx+b^2+c^2=0$$ After solving, you need to check whether the obtained $x$ values are sensible – they have to lie within 0 and $c$.

Solution 2:

Computing total area in two ways, we have $$ \frac{1}{b}(c+x) = \frac{1}{2}b(c-x) + a\sqrt{b^2+(c-x)^2} $$ and hence $$\frac{bx}{a} = \sqrt{b^2+(c-x)^2} $$ The rest of proof is same as that in Parcly Taxel's solution.

Solution 3:

Let's define an angle $\alpha$ which defines the line between the lines noted a and b. Then we have $x=a/\cos\alpha$ and $c=x+b\tan\alpha$. Using $x=\cos^{-1}(\alpha/x)$, $\tan\alpha=\sin\alpha/\cos\alpha$ and $\sin\cos^{-1}y=\sqrt{1-y^2}$, we get $c=x+b\sqrt{1-\alpha^2/x^2}/(a/x)$. Simplifying that we get Parcly Taxel's solution.

Solution 4:

In the diagram there is a parallelogram on the left, and a right triangle on the right, and I denote by $y$ the oblique segment which is both a side of the parallelogram and the hypotenuse of the triangle.

By computing the area of the parallelogram in two ways, we see: $$bx = ay$$ but $y$ is also easy to determine from Pythagoras applied to the right triangle, so plugging in: $$bx = a\sqrt{b^2+(c-x)^2}$$ which is the same equation all the other answer arrive at.