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

#### no_arg

##### Member
Suppose that PQR is an equilateral 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.

Last edited:

#### Drdusk

##### π
Moderator
I was that student btw

#### Drdusk

##### π
Moderator
I think it's because X must only be in the center for a triangle to form?

#### no_arg

##### Member
I think it's because X must only be in the center for a triangle to form?
No X can be any interior point. PX QX and RX need to be lifted out of the diagram to form another triangle.

#### fan96

##### 617 pages
No X can be any interior point. PX QX and RX need to be lifted out of the diagram to form another triangle.
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...)

#### ultra908

##### Member
hehe i have spoilers...

#### no_arg

##### 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...)
Given similarity, the solution cannot depend on the side length of the original triangle. Answer is disturbingly simple and the proof is stunning.

#### no_arg

##### Member
Both proofs a little clumsy

#### no_arg

##### Member
Simply rotate the entire diagram about P anticlockwise by sixty degrees and voila!

#### fan96

##### 617 pages
Given similarity, the solution cannot depend on the side length of the original triangle. Answer is disturbingly simple and the proof is stunning.
The quantity

$\bg_white \frac{k^2-RX^2}{2\cdot PX \cdot QX}$

can probably be simplified so as to remove $\bg_white k$.

The answer I gave holds numerically for all such equilateral and isosceles triangles formed and most likely for the rest of them too.

The other solutions are definitely much nicer though.

Both proofs a little clumsy
I think the first answer given is essentially the same as the one you posted, just a bit more direct.

#### no_arg

##### Member
The quantity

$\bg_white \frac{k^2-RX^2}{2\cdot PX \cdot QX}$

can probably be simplified so as to remove $\bg_white k$.

The answer I gave holds numerically for all such equilateral and isosceles triangles formed and most likely for the rest of them too.

The other solutions are definitely much nicer though.

I think the first answer given is essentially the same as the one you posted, just a bit more direct.
The use of congruence makes it less direct.