but I purposely posted this particular problem (with these particular values) because it does (i think) have a geometric solution. The particular relationships that arise out of these angles do 'tease' one's cognitive prowess !
ORIGINAL PROBLEM -
ABC is an isosceles triangle with angles A,B and C measuring 20,80 and 80 degrees respectively. Two rays BX and CY are drawn such that angle CBX measures 60 degrees and angle BCY measures 50 degrees. X and Y lie on sides AC and AB respectively.(i.e. the angles lie on the same side of side BC as the vertex A). Join XY. Find measure angle AXY.

I had it from a friend appearing for the Mathematics Olympiads(whose whereabouts i am not aware of right now :( ) six years ago. Here is wat i recall - (As you read ahead, take a look in the attached file)

CONSTRUCTION - Draw angle CBZ 20 degrees with Z lying on AC. Join YZ. Let us jot down what do we have now -
1. Triangle CBZ is isosceles. So, BC = BZ
2. Triangle BCY is isosceles. So, BC = BY.
1 and 2 give BY = BZ
3. Triangle BYZ is equilateral as BY = BZ and angle YBZ = 60 degrees. So, angle BYZ is 60 degrees.
So we have : So angle CYZ is 10 degrees AND BC=BY=BZ=YZ.

getting angle CYZ as 10 degrees was an intermediate step. Hereafter, he mentioned using one more construction of a circular arc.
Can you help me ?

Hi DeepG,

Using the same method I used in the thread
here,
I got a value of the angle as the solution to
sin(110-x)/sin(x) = sin(30)sin(60)/(sin(20)sin(40)), which gives x = 25.10437 degrees. Just as with the triangle problem in the other thread, this means that this angle is not one you can construct easily from the given angles (20,80,50,60,180). Therefore, I am forced to conclude that someone has misled you. This problem is NOT geometrically solvable with angles of 50 and 60 degrees, for exactly the same reason that the other problem is not solvable for 20 and 30 degrees. The trigonometric solution is not too hard (see above), but it is not the kind of solution you had hoped for.

Sorry for the bad news,

--Stuart Anderson

The angle I was trying to find is BXY, while you found AXY. If your answer is correct, then BXY should be 30 degrees, while I got approximately (but not exactly) 25 degrees. One of us is wrong!

I also have a question about your step 6. I drew the diagram according to your instructions, and I understand all the steps, and the picture I got makes sense. I also agree with all the other steps, but in step 6 I do not see how you know that the meeting point of the lines through R and R2 is at the center of the circle CBRR2. It is clear that RR2O is equilateral, and that O is on the vertical axis of symmetry of the circle, but I just do not see why it is at the center.

Thanks,

--Stuart Anderson

I am going to give you the reasoning behind I assumed O is the center of the circumference CBR2R, it is more a sort a quick check than a formal proof.

1) Angle RR2B equal 100 degrees since opposed angle in a isosceles trapezoid must add to 180 and angle in C equals 80.

2) As RR2O is 60 then OR2B is 40, since I assume OR2 and OB are the radius angle R2BO is also 40 so the central angle BOR2 is 100.

3) Same numbers for symmetrical isosceles triangle COR, (central angle COR equals 100). The above leads us to the triangle COB is also identical to the others OR2B and COR central angle (COB 100) . Angle ROR2 must be 60 as supposed by hipothesis and the important thing O is equidistant to R,R2, C and B. The equality of the three isosceles triangle get also validated taking under consideration the value of 50 degree for the angle BCY in the original set.

Of course this is not a rigorous proof. I believe the right way to do it'should be assuming a different point call it M as a center for the circumference CBR2R and then working for �reductio ab absurdum � so M must be O.

But Stuart� I am just an end user of math, .. do not ask me for more now I feel I got the flu.