Problem:
Let ABCD be a cyclic quadrilateral with AB=4,BC=5,CD=6, and DA=7. Let A1 and C1 be the feet of the perpendiculars from A and C, respectively, to line BD, and let B1 and D1 be the feet of the perpendiculars from B and D, respectively, to line AC. The perimeter of A1B1C1D1 is nm, where m and n are relatively prime positive integers. Find m+n.
Solution:
Let P denote the intersection point of diagonals AC and BD, and let θ be the acute angle formed by AC and BD. Because ∠DC1C=∠DD1C=90∘, it follows that CDC1D1 is cyclic, implying that ∠PD1C1=∠PDC and ∠PC1D1=∠PCD.
Let X be the reflection of B across the perpendicular bisector of diagonal AC. Then ABXC is an isosceles trapezoid, so A,B,X,C, and D lie on a circle. Because \overparen{A B}=\overparen{X C},
\angle X A D=\frac{\overparen{X C}+\overparen{C D}}{2}=\frac{\overparen{A B}+\overparen{C D}}{2}=\angle A P B=\theta
Similarly, ∠XCD=∠APD=180∘−θ. Applying the Law of Cosines to △XCD and △XAD gives
XD2=42+62+2⋅4⋅6cosθ=52+72−2⋅5⋅7cosθ
so cosθ=5911. Therefore the perimeter of A1B1C1D1 is 22⋅5911=59242. The requested sum is 242+59=301.
As in the first solution, the perimeter of A1B1C1D1 equals 22cosθ.
Note that the area of ABCD equals 2AC⋅BD⋅sinθ. On the other hand, by Brahmagupta's formula, area of cyclic quadrilateral ABCD equals
(s−a)(s−b)(s−c)(s−d)
where a,b,c,d are side lengths and s is the semiperimeter. In this case,
2AC⋅BD⋅sinθ=4⋅5⋅6⋅7
By Ptolemy's Theorem, AC⋅BD=4⋅6+5⋅7=59. Hence
sinθ=592⋅840
from which cosθ=5911 and the solution finishes as above.
