pariearth wrote:
In the following figure find angle ACB. Given C is the point of contact of two circles and A, B are the points of contact of the common tangent PP1 to the two circles.
(Find figure in the attched file)
(A) 60
(B) 90
(C) 120
(D) 89
(E) Cannot be determined
Dear
PariearthThis is a very hard Geometry problem, much harder than you will see on the GMAT. I'm pretty good with Geometry, and it took me several minutes to puzzle out this particular problem.
Attachment:
tangent circles tangent to line.JPG [ 26.62 KiB | Viewed 2202 times ]
In the diagram above, D and E are centers of the circle. Of course, AD = CD, so ACD is an isosceles triangle, and similarly, CE = BE, so BCE is an isosceles triangle as well. We will need Mr. Euclid's remarkable theorem, the
Isosceles Triangle Theorem.
Let angle DAC = p, and let the complement of p be q ----> p + q = 90, and 2p + 2q = 180.
In triangle ACD, by the Isosceles Triangle Theorem, angle DAC = angle ACD = p, and because the three angles of triangle ACD must sum to 180, we know angle D = 180 - 2p = 2q. The angles of triangle ACD are p & p & (2q).
Now, look at angle E. Segments AD and BE are parallel (they are both perpendicular to line AB, and perpendicular to the same thing means they're parallel). Because AD // BE, angles D & E are
same side interior angles, which must be supplementary. Since angle D = (2q), angle E = (2p).
We know angle (BCE) = angle (CBE), again by the ITT. Because the three angles of triangle BCE must sum to 180, we can deduce that angle (BCE) = angle (CBE) = q. The angles of triangle BCE are q & q & (2p).
Now, look at the three angles at point C. The three angles form a straight line, so their sum must be 180.
(angle ACD) + (angle ACB) + (angle BCE) = 180
p + (angle ACB) + q = 180
(angle ACB) + (p + q) = 180
(angle ACB) + 90 = 180
(angle ACB) = 90
Answer =
BDoes all this make sense?
Mike
_________________
Mike McGarry
Magoosh Test PrepEducation is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)