2
$\begingroup$

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 $x$.

enter image description here

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.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

Considering the right triangle BCD, one can write :

$$y=(3x-4)\tan(36°)\tag{1}$$

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

$$x=3.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.