Problem:
Find the area of rhombus ABCD given that the radii of the circles circumscribed around triangles ABD and ACD are 12.5 and 25 , respectively.
Solution:
Let O be the point of intersection of diagonals AC and BD, and E the point of intersection of AC and the circumcircle of β³ABD. Extend DB to meet the circumcircle of β³ACD at F. From the Power-of-a-Point Theorem, we have
AOβ OE=BOβ OD and DOβ OF=AOβ OC
Let AC=2m and BD=2n. Because AE is a diameter of the circumcircle of β³ABD, and DF is a diameter of the circumcircle of β³ACD, the above equalities can be rewritten as
m(25βm)=n2 and n(50βn)=m2
or
25m=m2+n2 and 50n=m2+n2
Therefore m=2n. It follows that 50n=5n2, so n=10 and m=20. Thus [ABCD]=(1/2)ACβ BD=2mn=400.
OR
Let R1β and R2β be the circumradii of triangles ABD and ACD, respectively. Because BO is the altitude to the hypotenuse of right β³ABE,AB2=AOβ AE. Similarly, in right β³DAF,AB2=DA2=DOβ DF, so AOβ AE=DOβ DF. Thus
DOAOβ=AEDFβ=R1βR2ββ=2
Also, from right β³ADE,2=DOAOβ=OEDOβ. Then
25=2R1β=AE=AO+OE=2β DO+21βDO=25βDO
and DO=10,AO=20, so [ABCD]=400.
OR
Let s be the length of a side of the rhombus, and let Ξ±=β BAC. Then AO=scosΞ±, and BO=ssinΞ±, so [ABCD]=4[ABO]=2s2sinΞ±cosΞ±=s2sin2Ξ±. Apply the Extended Law of Sines (In any β³ABC with AB=c,BC=a, CA=b, and circumradius R,sinAaβ=sinBbβ=sinCcβ=2R) in β³ABD and β³ACD to obtain s=2R1βsin(90ββΞ±)=sin2R1βcosΞ±, and s=2R2βsinΞ±. Thus tanΞ±=cosΞ±sinΞ±β=R2βR1ββ=21β. Also, s2=4R1βR2βcosΞ±sinΞ±=2R1βR2βsin2Ξ±. But sin2Ξ±=2β 5β1ββ 5β2β=54β, from which [ABCD]=2R1βR2βsin22Ξ±=2β 225ββ 25β 2516β=400β .
OR
Let AB=s,AO=m, and BO=n, and use the fact that the product of the lengths of the sides of a triangle is four times the product of its area and its circumradius to obtain 4[ABD]R1β=sβ sβ 2n and 4[ACD]R2β=sβ sβ 2m. Since [ABD]=[ACD], conclude that 21β=R2βR1ββ=mnβ, and proceed as above.
