<p>Well, so I'm a junior in high school and right now I have a B+ (92.34 ...arghhhh!!) in my AP Calc class.</p>
<p>Since I'm so close to getting an A (less than .2 away), my teacher gave me this problem to do for extra credit and said that I could get any help I wanted (thank god, or I would never be able to do this).</p>
<p>The text for the problem is on the bottom and the diagram is the one on the right (exercise 28).</p>
<p>So I would really really really really really greatly appreciate it if one of you guys could figure this out and tell me how to do it (eg. where to draw auxilary lines, steps in proof).</p>
<p>Lol, i dont even get why I get a geometry problem for AP Calc but w/e.</p>
<p>So any help greatly appreciated! If you have any questions PM me! Please try to get this by tonight! </p>
<p>If we draw the center O of the circle, then since ABC is an isosceles triangle, AO (extended) will be the altitude from C to AB, therefore AO is perpendicular to AB. Because AO goes through the center of the circle, it is perpendicular to CD. Therefore CD is parallel to AB. Therefore ABF and DCF have the same angles. Therefore FD/CD=AF/AB =6/3 = 2. So 10+x = 2y, implying y = 5 + x/2. We also get the equation x(x+16) = y^2 from the Power of a Point theorem on point D. Thus x^2+16x = (5+x/2)^2 = 25 + 5x + x^2/4 which implies 3x^2/4 + 11x - 25 = 0. By the quadratic formula, x = (-11 + sqrt(11^2 + 100*3/4))/(3/2) = (-11 + 14)/(3/2) = 2.</p>
<p>This doesn’t seem right. AO is only perpendicular to AB if AB was a tangent to the circle. This is not true. </p>
<p>Incidentally, if you assume that angle ABF is a 90 degree angle and proceed using the fact that the two triangles ABF and FCD are similar triangles, you come out with the equation 10 + x = 2y. Since angle ABF is not 90 degrees, this equation is probably not valid. It seems cruel to have a triangle in there that looks like a right triangle and has two sides that make it look like a 30-60-90 triangle, but ABF cannot be a right triangle.</p>
<p>Aren’t there equations for sin (x) that don’t assume you are dealing with a right triangle. I forget what they are exactly. The fact that angle AFB and angle CFD are the same is undoubtedly key to solving this problem. If you can set sin (AFB) = sin (CFD), then maybe you can solve it.</p>
<p>OK, google the “Law of Cosines”. It deals with the cosine of an angle in an oblique triangle (not a right triangle) and expresses it in terms of the lengths of the three sides. It looks like the Pythagorean thereom with an added term.</p>
<p>I won’t try to write it here because its to difficult to describe what the variables refer to.</p>
<p>hey guys, thanks for the replies. so i get all the algebra part but am still confused with how we’re getting the triangles similar…could you clarify it and tell me what thrms/corolaries you used?</p>
<p>Draw the center O of the circle. Then draw the perpendicular bisector l of AB. Because OA = OB, l passes through O. Because CA = CB, l passes through C. Thus, l contains the radius OC.</p>
<p>Since radii are perpendicular to tangents at the point of tangency, OC is perpendicular to CD. Since l contains the segment OC, l is perpendicular to CD. Since l is the perpendicular bisector of AB, l is perpendicular to AB.</p>
<p>Because l is perpendicular to both CD and AB, AB is parallel to CD.</p>
<p>Therefore <ABC and <BCD are “alternate interior angles” so they have equal measure. Also, <BAD and <CDA are “alternate interior angles” so they have equal measure.</p>
<p>Therefore, since ABF shares two angle measures with DCF, it is similar to it.</p>
