AnExtremelyLongName

Hint: AT = AM because ED is a perpendicular bisector. Whenever a chord intersects and forms a right angle, then one of them is a perpendicular bisector.

Can you at least give some clue on how you got that answer?

Are you allowed to use a calculator?

Hello madyl, this is my insane solution that is probably wrong.

Extend BD to point that we call \(L\) in such that \(BL=BE=EL\), forming equilateral triangle \(BEL\). Copy triangle ABC to the other side of CED to make the entire picture symmetrical.

Since ABC and the other triangle are equilateral and congruent, we know BC = DL. We also know that Angle B = Angle L. Since we made the picture symmetrical, we know that BE = LE. Because of SAS, we know that triangles EBC is congruent to ELD. Because of CPCTC, we know that (B) CE = ED.

 kind of weird solution I came up with

For example: 123 is the smallest 3 digit integer with digits strictly increasing left to right.

We list smallest numbers like cases in the example:

1 (does not count)

12

123

1234

12345

123456

1234567

12345678

123456789

Notice how for each of these cases. The number with \(n\) digits has \(10-n\) cases that fit the problem. Starting with the two digit numbers, the number of cases that fit the problem is 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = \(\boxed{36}\)

I'm not sure if this is correct, but it is my attempt on the problem.

Nice!  I like that strategy!

You missed some latex!

The largest digit possible is 123456789. There is a limit.

5 step solution

1. Divide AOB in half and use right triangle trigonmetry to find the base and height. (Alternative method, you can use law of cosines to find AB then proceed with Heron's formula)

2. Knowing the base and height, you can find the area of triangle AOB.

3. Find the area of the major portion of the circle that contains major arc AB

4. Add the area of the major portion and triangle together. Subtract that from the total circle area.

5. Your result is your answer.

Extend CE and extend AB, let us call the point of intersection \(L\)

 - Because of vertical angles angle \(LCB\) is 55 degrees.

 - Angle \(CBL\) is 180 - 120 = 60 degrees.

Since all angles in a triangle add up to 180 degrees, angle \(CLB\) = 65 degrees. Because of alternate interior angles, the supplement of CLB is congruent to CEF.

Therefore, CEF = 180 - 65 = \(\boxed{115}\) degrees.