Maths Olympiad Prep

Track / Stage 4 / 197 of 340 #457 of 1964

Problem 457

AMC 12 late, AIME early
Geometry Difficulty 4.8 Find the answer

3. Given a right trapezoid ABCDA B C D with side lengths AB=2,BC=CD=10,AD=6A B=2, B C=C D=10, A D=6, a circle is drawn through points BB and DD, intersecting the extension of BAB A at point EE and the extension of CBC B at point FF. Then the value of BEBFB E-B F is \qquad

A number or a short expression. Spacing, $ signs and \frac vs / are all fine.

Official solution

3.4 .

As shown in Figure 3, extend CDC D to intersect O\odot O at point GG. Let the midpoints of BEB E and DGD G be MM and NN, respectively. It is easy to see that AM=DNA M = D N. Since BC=CD=10B C = C D = 10, by the secant theorem, it is easy to prove that
BF=DG. Therefore, BEBF=BEDG=2(BMDN)=2(BMAM)=2AB=4. \begin{aligned} B F = & D G. \text{ Therefore, } \\ & B E - B F = B E - D G = 2(B M - D N) \\ & = 2(B M - A M) = 2 A B = 4 . \end{aligned}

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.