only 1% of people can solve it

@Gautham_A said in #20:

Me when no options were given:
media1.tenor.com/m/-i9mRdfpygYAAAAd/squidward-throw-away.gif
LOL

@Thalassokrator @Noflaps
like I said, here was very quick solution for it.
you can calculate all angles in SQPE and prove that both angle opposites are equal 180 degrees.
with means it can be cirlce described in SQPE quadrangle. where diameter is 6;
let's define S2 with is center of circle described in SQPE quadrangle.
|S2Q| = |S2P| = 3; and given a law of circles,|S2QP| = 90 degres (because of |EPQ| = 45 degrees).
and using pitagoras theory. 3^2 + 3^2 = |PQ|^2;
|PQ| = 3√2
imagine having that question in highschool exam (in poland). for only 3 points of 50. it is one of the hardest problems in highschool tho(at extended math only).

@Damkiller25
is it 6 under root 2
6√2

edit:oh okay , now see your forum #22...
okay my answer is wrong, well thanks for the solution

ofcourse the answer is ................................................................
4.2348372838737477378274874838483793220384743748387482824878273784824847248247897427489274892798274892742749832748923743289748293749824729837489237489274892748927482974892734892748923748237894 and sooo on

#1
Omg, which grade problem is that!!?!

grade 2, that's who

@Damkiller25 said in #22:

like I said, here was very quick solution for it.
you can calculate all angles in SQPE and prove that both angle opposites are equal 180 degrees.

Oh yes, indeed the opposite angles in the cyclic quadrilateral sum to 180º due to the inscribed angle theorem. Very clever!
Notice the proof without words for the supplementary angles in a quadrilateral in the last figure on this page in particular:
https://en.wikipedia.org/wiki/Inscribed_angle

with means it can be cirlce described in SQPE quadrangle. where diameter is 6;

True, SQEP can be inscribed in a circle (as in the inscribed angle theorem above) and since SE is the radius of the larger circle 6 this smaller circle must have radius 3.

let's define S2 with is center of circle described in SQPE quadrangle.

So S2 is the centre of the smaller circle in which the quadrilateral SQEP is inscribed, I take it.

|S2Q| = |S2P| = 3; and given a law of circles,|S2QP| = 90 degres (because of |EPQ| = 45 degrees).

I believe you are thinking of angles ∠QS2P = 90º and ∠PEQ = 45º respectively (https://en.wikipedia.org/wiki/Angle#Identifying_angles).
Those again follow from the inscribed angle theorem. You are describing cases 2) and 4) as defined in my previous post where I called ∠PEQ = Ω. In cases 1) and 3) ∠PEQ is 135º, but the inscribed angle theorem can still be applied if we instead use the central angle of the large circle ∠PSQ = 45º which becomes the inscribed angle of the small circle. This again yields ∠QS2P = 90º as the central angle of the small circle.

So in total, you have constructed an isosceles right triangle with sides |S2Q| = |S2P| = 3 and |PQ|=?. The fact that in any case ∠QS2P = 90º obviates the necessity to use the law of cosines and allows you to simply use Pythagoras' theorem! Very neat!

and using pitagoras theory. 3^2 + 3^2 = |PQ|^2;
|PQ| = 3√2
imagine having that question in highschool exam (in poland). for only 3 points of 50. it is one of the hardest problems in highschool tho(at extended math only).

That's indeed a tricky problem for high school. Hope you did well on your exam! I liked working through your solution, thank you!
Perhaps I should have spotted the possibility of using the inscribed angle theorem, alas I didn't even consider it. It's VERY useful and comes up quite frequently in such problems. The first one that comes to mind is this extremely beautiful problem:
https://www.youtube.com/watch?v=d-o3eB9sfls&t=762s
So much fun that infinite series can be related to elemental geometry and the physics of light intensity in 3D space in such a fundamental way. Truly awe-inspiring!

Caution: incoming dad joke.

It's so frustrating that pi is always popping up in unexpected places in the universe -- but when I get an urge for pie I am so seldom offered any. . . .

call @ForumMathSolver

A simpler approach would be to use cartesian coordinates. Assume the circle has its center at (0, 0). With radius 6 units, we would get its equation as (x - 0)^2 + (y - 0)^2 = r^2 or x^2 + y^2 = 36. Assuming one of the diameters (say AB) to be the x-axis, we could get the equation of the other diameter at a inclination of 45 degrees to be x = y.

Assume the point E to be (r cos theta, r sin theta) = (6 cos theta, 6 sin theta), we could get the equations of the two projections EP and EQ, and, consequently, the coordinates of the points P and Q, in terms of theta. Then we could apply distance formula to calculate the length of PQ. Since sin^2 theta + cos^2 theta = 1, they should cancel each other out.