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2018 AMC 12A Problem 20

Problem 20 of 25HarderGeometry

Triangle ABCABC is an isosceles right triangle with AB=AC=3.AB = AC = 3. Let MM be the midpoint of hypotenuse BC‾.\overline{BC}. Points II and EE lie on sides AC‾\overline{AC} and AB‾,\overline{AB}, respectively, so that AI>AEAI \gt AE and AIMEAIME is a cyclic quadrilateral. Given that triangle EMIEMI has area 2,2, the length CICI can be written as a−bc,\tfrac{a - \sqrt{b}}{c}, where a,a, b,b, and cc are positive integers and bb is not divisible by the square of any prime. What is the value of a+b+c?a + b + c?

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Solution

Since △ABC\triangle ABC is an isosceles right triangle, CM=BM=322CM = BM = \tfrac32 \sqrt2 and the base angles at B,CB, C are 45∘.45^\circ. As AIMEAIME is cyclic with right angle at A,A, angle ∠IME=90∘.\angle IME = 90^\circ. Let x=CIx = CI and y=BE.y = BE. By the Law of Cosines in △MCI,\triangle MCI, IM2=x2+92−2⋅x⋅322⋅cos⁡45∘=x2−3x+92, \begin{aligned} IM^2 &= x^2 + \tfrac92 \\ &\quad {}- 2 \cdot x \cdot \tfrac32\sqrt2 \cdot \cos 45^\circ \\ &= x^2 - 3x + \tfrac92, \end{aligned} and similarly ME2=y2−3y+92.ME^2 = y^2 - 3y + \tfrac92. The Pythagorean Theorem in right triangles EMIEMI and IAEIAE gives IM2+ME2IM^2 + ME^2 =(3−x)2= (3-x)^2 +(3−y)2,+ (3-y)^2, which simplifies to x+y=3.x + y = 3. The area condition 12IM⋅ME=2\tfrac12 IM \cdot ME = 2 means IM2⋅ME2=16.IM^2 \cdot ME^2 = 16. Substituting y=3−xy = 3 - x makes ME2=x2−3x+92=IM2,ME^2 = x^2 - 3x + \tfrac92 = IM^2, so (x2−3x+92)2=16,\left(x^2 - 3x + \tfrac92\right)^2 = 16, hence x2−3x+92=4,x^2 - 3x + \tfrac92 = 4, i.e. x2−3x+12=0.x^2 - 3x + \tfrac12 = 0. Since AI>AEAI \gt AE forces y>x,y \gt x, we take the smaller root x=3−72.x = \tfrac{3 - \sqrt{7}}{2}. Then a+b+c=3+7+2=12.a + b + c = 3 + 7 + 2 = 12. Thus, the correct answer is D.
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Tagged: cyclic quadrilateral · law of cosines · triangle area

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