Problem:
Let ABCD be a convex quadrilateral with BC=2 and CD=6. Suppose that the centroids of β³ABC,β³BCD, and β³ACD form the vertices of an equilateral triangle. What is the maximum possible value of the area of ABCD ?
Answer Choices:
A. 27
B. 163β
C. 12+103β
D. 9+123β
E. 30 Solution:
Place an origin at A, and assign position vectors of B=pβ and D=qβ. Since AB is not parallel to AD, vectors pβ and qβ are linearly independent, so we can write C=mpβ+nqβ for some constants m and n. Now, recall that the centroid of a triangle β³XYZ has position vector 31β(x+yβ+z).
Thus the centroid of β³ABC is g1β=31β(m+1)pβ+31βnqβ; the centroid of β³BCD is g2β=31β(m+1)pβ+31β(n+1)qβ; and the centroid of β³ACD is g3β=31βmpβ+31β(n+1)qβ.
Hence G1βG2ββ=31βqβ,G2βG3ββ=β31βpβ, and G3βG1ββ=31βpββ31βqβ. For β³G1βG2βG3β to be equilateral, we need β£β£β£β£βG1βG2βββ£β£β£β£β=β£β£β£β£βG2βG3βββ£β£β£β£βββ£pββ£=β£qββ£βAB=AD. Further, β£β£β£β£βG1βG2βββ£β£β£β£β=β£β£β£β£βG1βG3βββ£β£β£β£βββ£pββ£=β£pββqββ£=BD. Hence we have AB=AD=BD, so β³ABD is equilateral.
Now let the side length of β³ABD be k, and let β BCD=ΞΈ. By the Law of Cosines in β³BCD, we have k2=22+62β2β 2β 6β cosΞΈ=40β24cosΞΈ. Since β³ABD is equilateral, its area is 43ββk2=103ββ63βcosΞΈ, while the area of β³BCD is 21ββ 2β 6β sinΞΈ=6sinΞΈ. Thus the total area of ABCD is 103β+6(sinΞΈβ3βcosΞΈ)=103β+12(21βsinΞΈβ23ββcosΞΈ)=103β+12sin(ΞΈβ60β), where in the last step we used the subtraction formula for sin. Alternatively, we can use calculus to find the local maximum. Observe that sin(ΞΈβ60β) has maximum value 1 when e.g. ΞΈ=150β, which is a valid configuration, so the maximum area is 103β+12(1)=(C)12+103ββ.
OR
Let G1β,G2β,G3β be the centroids of ABC,BCD, and CDA respectively, and let M be the midpoint of BC. A,G1β, and M are collinear due to well-known properties of the centroid. Likewise, D,G2β, and M are collinear as well. Because (as is also well-known) 3MG1β=AM and 3MG2β=DM, we have β³MG1βG2ββΌβ³MAD. This implies that AD is parallel to G1βG2β, and in terms of lengths, AD=3G1βG2β. (SAS Similarity)
We can apply the same argument to the pair of triangles β³BCD and β³ACD, concluding that AB is parallel to G2βG3β and AB=3G2βG3β. Because 3G1βG2β=3G2βG3β (due to the triangle being equilateral), AB=AD, and the pair of parallel lines preserve the 60β angle, meaning β BAD=60β. Therefore β³BAD is equilateral.
At this point, we can finish as in Solution 1, or, to avoid using trigonometry, we can continue as follows:
Let BD=2x, where 2<x<4 due to the Triangle Inequality in β³BCD. By breaking the quadrilateral into β³ABD and β³BCD, we can create an expression for the area of ABCD. We use the formula for the area of an equilateral triangle given its side length to find the area of β³ABD and Heron's formula to find the area of β³BCD.
After simplifying,
[ABCD]=x23β+36β(x2β10)2β
Substituting k=x2β10, the expression becomes
[A B C D]=k\sqrt{3}+\sqrt{36-k^{2}}+10\sqrt
We can ignore the 103β for now and focus on k3β+36βk2β.
The RHS simplifies to 122, meaning the maximum value of k3β+36βk2β is 12 .
Thus the maximum possible area of ABCD is (C)12+103ββ.
OR
Let A,B,C, and D correspond to the complex numbers a,b,c, and d, respectively. Then, the complex representations of the centroids are (a+b+c)/3,(b+c+d)/3, and (a+c+d)/3. The pairwise distances between the centroids are β£(dβa)/3β£,β£(bβa)/3β£, and β£(bβd)/3β£, all equal. Thus, β£(bβa)/3β£=β£(dβa)/3β£=β£(bβd)/3β£, so β£(bβa)β£=β£(dβa)β£=β£(bβd)β£. Hence, β³DBA is equilateral.
By the Law of Cosines, [ABCD]=[ABD]+[BCD]=4(22+62β2β 2β 6cos(β BCD)β)2β 3ββ+1/2β 2β 6sin(β BCD).
[ABCD]=103β+6(sinβ BCDβ3βcos(β BCD))=103β+12sin(β BCDβ60β)β€12+103β. Thus, the maximum possible area of ABCD is (C)12+103ββ.