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2013 AMC 12B Problem 19

Problem 19 of 25HarderGeometry

In triangle ABC,ABC, AB=13,AB = 13, BC=14,BC = 14, and CA=15.CA = 15. Distinct points D,D, E,E, and FF lie on segments BC,BC, CA,CA, and DE,DE, respectively, such that AD⊥BC,AD \perp BC, DE⊥AC,DE \perp AC, and AF⊥BF.AF \perp BF. The length of segment DFDF can be written as mn,\dfrac{m}{n}, where mm and nn are relatively prime positive integers. What is m+n?m + n?

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Solution

The altitude from AA to BCBC gives BD=5,BD = 5, CD=9,CD = 9, AD=12.AD = 12. Because DE⊥AC,DE \perp AC, triangle AED∼ADC,AED \sim ADC, giving DE=365DE = \tfrac{36}{5} and AE=485.AE = \tfrac{48}{5}. Since ∠AFB=∠ADB=90∘,\angle AFB = \angle ADB = 90^\circ, quadrilateral ABDFABDF is cyclic, so ∠ABD=∠AFE,\angle ABD = \angle AFE, making right triangles ABDABD and AFEAFE similar: FE5=48512,\dfrac{FE}{5} = \dfrac{\frac{48}{5}}{12}, so FE=4.FE = 4. Hence DF=DE−FEDF = DE - FE =365−4= \tfrac{36}{5} - 4 =165,= \tfrac{16}{5}, and m+n=21.m + n = 21. Thus, the correct answer is B.

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Concepts: cyclic quadrilateral · similarity · altitude

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.