This problem was asked from 8th and 9th graders in a contest.

You have an quadrilateral $ABCD$ with $AB=4x-y$, $BC=3x-4$, $CD=y$, and $DA=5x-2$. You draw a line $BD$ which makes a right triangle with $\angle CBD=36°$ and in the triangle $ABD$ $\angle DBA=126°$. Solve for $x$.

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I was able to solve this with a calculator but this problem is supposed to be solved without one. I tried to find an answer to this problem online but I couldn't find one, probably because my math vocab in English isn't that great.


Nice question!

Reflect the upper left triangle in the diagonal and rotate the lower triangle about the mid-point of the diagonal. (Delete the original triangles.)

You should now be able to see a Pythagorean triangle with sides $3x-4,4x,5x-2$.

Then $x=3$ gives the solution and the Pythagorean triangle is a $(5,12,13)$ triangle.


Considering the right triangle BCD, one can write :


Let us apply to the other triangle ABD the law of cosines:

$$AD^2=AB^2+BD^2-2 AB.BD \cos(126°)$$

$$=AB^2+\underbrace{(BC^2+CD^2)}_{\text{Pythagoras}}-2 AB.\dfrac{BC}{\cos(36°)} (-\sin(36°))$$

$$(5x-2)^2=(4x-y)^2+y^2+(3x-4)^2+2 (4x-y).\underbrace{(3x-4)\tan(36°)}_{= \ y \ \ \text{using} \ (1)} \ \ \ \ \ (2)$$

Expanding (2), which is an identity, one obtains $4x-12=0$, imposing value



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