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2021 AMC 12B Problem 24

Problem 24 of 25HarderGeometry

Let ABCDABCD be a parallelogram with area 15.15. Points PP and QQ are the projections of AA and C,C, respectively, onto the line BD;BD; and points RR and SS are the projections of BB and D,D, respectively, onto the line AC.AC. See the figure, which also shows the relative locations of these points. Suppose PQ=6PQ=6 and RS=8,RS=8, and let dd denote the length of BD‾,\overline{BD}, the longer diagonal of ABCD.ABCD. Then d2d^2 can be written in the form m+np,m+n\sqrt p, where m,m, n,n, and pp are positive integers and pp is not divisible by the square of any prime. What is m+n+p?m+n+p?

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Solution

Let the diagonals meet at OO at angle θ.\theta. The feet of the perpendiculars from AA and CC to BDBD are symmetric about O,O, so PQ=ACcos⁡θ=6;PQ=AC\cos\theta=6; likewise RS=BDcos⁡θ=8.RS=BD\cos\theta=8. The parallelogram’s area is 12⋅AC⋅BDsin⁡θ=15,\tfrac12\cdot AC\cdot BD\sin\theta=15, so AC⋅BDsin⁡θ=30.AC\cdot BD\sin\theta=30. Then 48sin⁡θcos⁡2θ=30,\dfrac{48\sin\theta}{\cos^2\theta}=30, giving sin⁡θcos⁡2θ=58.\dfrac{\sin\theta}{\cos^2\theta}=\dfrac58. Writing s=sin⁡θ,s=\sin\theta, 8s=5(1−s2)8s=5(1-s^2) gives s=−4+415,s=\dfrac{-4+\sqrt{41}}{5}, so cos⁡2θ=1−s2=8(41−4)25.\cos^2\theta=1-s^2=\dfrac{8(\sqrt{41}-4)}{25}. Then d2=BD2d^2=BD^2 =64cos⁡2θ=\dfrac{64}{\cos^2\theta} =8(41+4)=8(\sqrt{41}+4) =32+841,=32+8\sqrt{41}, so m+n+p=32+8+41=81.m+n+p=32+8+41=81. Thus, the correct answer is A.
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Tagged: parallelogram · trigonometry · area

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