I get answer wrong.

[granite grove]

The triangles are similar at first sight, congruency may be a little trickier to assert. To prove congruency maybe prove that the octagon is regular (PW=WV=VU=...)

[prime sage]

ok they arent congruent

if the octagons regular, the triangles angles come 45 which is not possible seeing the fig

[granite grove]

hmm, can you elaborate?

I see what you mean now. Maybe the octagon still is equilateral, but this would be harder to prove than if it just were regular. So we may need to think about other way to prove congruency.

At the very least I confirmed in geogebra that the triangles are pretty much congruent. But that's of course not enough.

[frail musk]

It's easy to see that if $AP=x$, then $AQ=1-\frac{45}{101}-x$

[lapis steppe]

Civil Service Pigeon

[frail musk]

Now, you can apply the Pythagorean theorem on triangle APQ

[prime sage]

I did that but

the calculation becones too long and if u stick with it

Two real positive roots come

[frail musk]

They both yield the same result

[frail musk]

By symmetry, if $x$ is a root, then so is $\frac{56}{101}-x$

[lapis steppe]

Civil Service Pigeon

[frail musk]

Let $y=\frac{56}{101}-x$. Then, $$\left(\frac{45}{101} \right)^2=x^2+y^2$$ $$x+y=\frac{56}{101}$$ Then, note the required area is $$1-2xy=1-[(x+y)^2-(x^2+y^2)]$$ at which point you can apply difference of squares

[prime sage]

thanks

i got it

90/101 is the answer

[lapis steppe]

Civil Service Pigeon