# Hardest geometry question in history answered by student trivially............ How? (1 Viewer)

#### idkkdi

##### Well-Known Member
There are some things beyond my power

Question Mark 1:30
i see you @Eagle Mum. tell your boy to try this 50-60-70 q. the 60 one is straight forward enough 50-60-70 should be aimo level or higher and im too rusty.

#### idkkdi

##### Well-Known Member
50-60-70 q proposed by the green-grey icon person. the equilateral one is easier.

#### idkkdi

##### Well-Known Member
u going to try? lol.

FYI:

Suppose that PQR is a 50-60-70 (respectively) triangle with an interior point X. Let angle PXR = s and angle QXR = t.
Find (in terms of s and t) the three angles of any triangle with side lengths equal to PX, QX and RX.

#### idkkdi

##### Well-Known Member
Must the angles be expressed only in terms of $\bg_white s$ and $\bg_white t$?

The best I could do is

$\bg_white s' = \arccos\left( \cos s + \frac{k^2-RX^2}{2\cdot PX \cdot QX}\right),$

$\bg_white t' = \arccos\left( \cos t + \frac{k^2-PX^2}{2\cdot QX \cdot RX}\right),$

where $\bg_white k$ is the side length of the equilateral triangle and $\bg_white s', \, t'$ are the angles of the newly formed triangle.

(Hopefully that's correct...)
interested in seeing how you got here.

#### Etho_x

##### Well-Known Member
If that was a question on one of my exam papers

My response: Idk you find em for me