A good geometry question.?

Question

In the figure below, 'O' is the center of the circle and there are three equal angles named 'alpha'. Find 'alpha'

http://i21.tinypic.com/nzh8pl.jpg

Answer 1

The answer is 50 degrees.
The geometry will lead you to an indeterminate problem where
45 < α < 60
Then you have to combine that with some really tiresome trigonometry, making use of the various isosceles triangles in the problem.

Angles:
BOC = DOE = 2α
DEA = BDC = BOD = EOC = 180 - 2α
DEO = OCE = 180 - 3α
OEC = 5α - 180
DBO = 4α - 180

For all these angles to be positive
45 < α < 60 NOT INCLUSIVE

Now applying the sine and cosine rules to various triangles:
Let OB = OC = y = radius
BOD is isosceles, therefore BD = y
OD = -2y*cos(2α)
Note that cos(2α) = -ve
Let DE = AE = x
OE = x/(2cosα)
AD = 2xcosα
BC = 2ysinα

Apply the sine rule to triangle ABE:
x/sin(4α - 180) = [y + x/(2cosα)]/sinα
Simplify to get
x*[sin(2α) + sin(4α)] = -2ycosα*sin(4α) …(1)

Apply the sine rule to triangle BEC:
2ysinα/sin(5α - 180) = [y + x/(2cosα)] / sin(270 - 4α)
Simplify to get
-2ycosα*[sin(5α) – 2sinα*cos(4α)] = xsin(5α) … (2)

Combine eqns 1 and 2 to get:
[sin(2α)+sin(4α)]* [sin(5α)–2sinα*cos(4α)] = sin(4α)*sin(5α)
which simplifies to
sin(α) + sin(5α) – sin(7α) = 0 … (3)

α = 50 is the only solution to eqn 3 in range.

Answer 2

Hick Ninja: You've swapped D and E throughout your answer, except in deriving equation 4. To be consistent with the rest of your answer it should be 180 - (180 - 2*alpha) = 2*alpha, which obviously won't get you anywhere by setting it equal to equation 3.

I agree with sahsjing: we can characterise α as between 45° and 60°, but I don't think there is a single solution. To see how we get this, note the following:
Angle BOC is twice angle BAC, so is 2α; so the vertically opposite angle DOE is also 2α. It follows that angle OED is 180°-3α. Hence we must have 3α < 180°, i.e. α < 60°.

Also, angle BOD must be 180°-2α, and so must angle BDO. So angle DBO must be 180° - 2(180°-2α) = 4α - 180°. It follows that 4α > 180°, i.e. α > 45°.

In triangle EOC we have angles of 5α-180° for OEC, 180°-2α for EOC and 180°-3α for ECO. These don't give us any new limits on α.

As far as I can see it is possible to construct a solution for any α in the range 45° to 60°, but I haven't formally proved it yet.

Answer 3

it looks like it'd be 60 degrees

if the top triangle has two a's in it and the a's are all equal to each other, then there's only two possibility: it's an equilateral triangle, or it's an isocoles triangle (however you spell it)

so then look at the two that are next to each other, the three degrees together have to form 180 degrees, right? therefore, a could only be 60 (from what i can tell)

Answer 4

You can check that α is between 45 and 60 degrees but not equal to either 45 or 60 degrees.

Answer 5

Assuming AEB and ADC are straight lines.

Statement : Reason
1. angle AEC = 2*alpha : add angles AED and CED
2. angle ACE = 180 - 3*alpha : angles in triangle ACE must add to 180.
3. angle CDE = 2*alpha : angles in triangle CDE must add to 180.
4. angle CDE = 180 - alpha : angles ADE + CDE must sum to 180 because they're on one side of a straight line.

Then just set equation 3 and 4 equal to each other:
2*alpha = 180 - alpha
3*alpha = 180
alpha = 60 degrees

Answer 6

I dont get it...