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

[no_arg]

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.

 

[Drdusk]

I was that student btw

 

[Drdusk]

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

 

[no_arg]

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]

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 and ?

The best I could do is





where is the side length of the equilateral triangle and are the angles of the newly formed triangle.

(Hopefully that's correct...)

 

[ultra908]

hehe i have spoilers...

 

[no_arg]

Must the angles be expressed only in terms of and ?

The best I could do is





where is the side length of the equilateral triangle and 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.

 

[quickoats]

Spoiler

Random point inside an equilateral triangle

Take any equilateral triangle and pick a random point inside the triangle. Draw from each vertex a line to the random point. Two of the three angles at the point are known let's say $x$,$y$. If the

math.stackexchange.com

 

[no_arg]

Spoiler

Random point inside an equilateral triangle

Take any equilateral triangle and pick a random point inside the triangle. Draw from each vertex a line to the random point. Two of the three angles at the point are known let's say $x$,$y$. If the

math.stackexchange.com

Both proofs a little clumsy

 

[no_arg]

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

 

[fan96]

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



can probably be simplified so as to remove .

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]

The quantity



can probably be simplified so as to remove .

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.

 

[jyu]

(s-60), (t-60), (300-s-t)

 

[jyu]

by the way, I was that student

 

[jyu]

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]

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.

so is anyone going to solve this?

 

[idkkdi]

so is anyone going to solve this?

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.

Can someone please solve this lol.

 

[Velocifire]

 

[idkkdi]

View attachment 29800

hit up jesus and tell him to solve the variant question.

 

[Velocifire]

hit up jesus and tell him to solve the variant question.

There are some things beyond my power

Question Mark 1:30
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