Problem:
Let ABCD be a cyclic quadrilateral with AB=4,BC=5,CD=6, and DA=7. Let A1β and C1β be the feet of the perpendiculars from A and C, respectively, to line BD, and let B1β and D1β be the feet of the perpendiculars from B and D, respectively, to line AC. The perimeter of A1βB1βC1βD1β is nmβ, 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 ΞΈ be the acute angle formed by AC and BD. Because β DC1βC=β DD1βC=90β, it follows that CDC1βD1β is cyclic, implying that β PD1βC1β=β PDC and β PC1βD1β=β 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, β XCD=β APD=180ββΞΈ. Applying the Law of Cosines to β³XCD and β³XAD gives
XD2=42+62+2β 4β 6cosΞΈ=52+72β2β 5β 7cosΞΈ
so cosΞΈ=5911β. Therefore the perimeter of A1βB1βC1βD1β is 22β 5911β=59242β. The requested sum is 242+59=301.
As in the first solution, the perimeter of A1βB1βC1βD1β equals 22cosΞΈ.
Note that the area of ABCD equals 2ACβ BDββ sinΞΈ. On the other hand, by Brahmagupta's formula, area of cyclic quadrilateral ABCD equals
(sβa)(sβb)(sβc)(sβd)β
where a,b,c,d are side lengths and s is the semiperimeter. In this case,
2ACβ BDββ sinΞΈ=4β 5β 6β 7β
By Ptolemy's Theorem, ACβ BD=4β 6+5β 7=59. Hence
sinΞΈ=592β 840ββ
from which cosΞΈ=5911β and the solution finishes as above.
