Problem:
Let ABCD be a convex quadrilateral with BC=2 and CD=6. Suppose that the centroids of △ABC,△BCD, and △ACD form the vertices of an equilateral triangle. What is the maximum possible value of the area of ABCD ?
Answer Choices:
A. 27
B. 163
C. 12+103
D. 9+123
E. 30 Solution:
Place an origin at A, and assign position vectors of B=p and D=q. Since AB is not parallel to AD, vectors p and q are linearly independent, so we can write C=mp+nq for some constants m and n. Now, recall that the centroid of a triangle △XYZ has position vector 31(x+y+z).
Thus the centroid of △ABC is g1=31(m+1)p+31nq; the centroid of △BCD is g2=31(m+1)p+31(n+1)q; and the centroid of △ACD is g3=31mp+31(n+1)q.
Hence G1G2=31q,G2G3=−31p, and G3G1=31p−31q. For △G1G2G3 to be equilateral, we need ∣∣∣∣G1G2∣∣∣∣=∣∣∣∣G2G3∣∣∣∣⇒∣p∣=∣q∣⇒AB=AD. Further, ∣∣∣∣G1G2∣∣∣∣=∣∣∣∣G1G3∣∣∣∣⇒∣p∣=∣p−q∣=BD. Hence we have AB=AD=BD, so △ABD is equilateral.
Now let the side length of △ABD be k, and let ∠BCD=θ. By the Law of Cosines in △BCD, we have k2=22+62−2⋅2⋅6⋅cosθ=40−24cosθ. Since △ABD is equilateral, its area is 43k2=103−63cosθ, while the area of △BCD is 21⋅2⋅6⋅sinθ=6sinθ. Thus the total area of ABCD is 103+6(sinθ−3cosθ)=103+12(21sinθ−23cosθ)=103+12sin(θ−60∘), 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(θ−60∘) has maximum value 1 when e.g. θ=150∘, which is a valid configuration, so the maximum area is 103+12(1)=(C)12+103.
OR
Let G1,G2,G3 be the centroids of ABC,BCD, and CDA respectively, and let M be the midpoint of BC. A,G1, and M are collinear due to well-known properties of the centroid. Likewise, D,G2, and M are collinear as well. Because (as is also well-known) 3MG1=AM and 3MG2=DM, we have △MG1G2∼△MAD. This implies that AD is parallel to G1G2, and in terms of lengths, AD=3G1G2. (SAS Similarity)
We can apply the same argument to the pair of triangles △BCD and △ACD, concluding that AB is parallel to G2G3 and AB=3G2G3. Because 3G1G2=3G2G3 (due to the triangle being equilateral), AB=AD, and the pair of parallel lines preserve the 60∘ angle, meaning ∠BAD=60∘. Therefore △BAD 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 △BCD. By breaking the quadrilateral into △ABD and △BCD, 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 △ABD and Heron's formula to find the area of △BCD.
After simplifying,
[ABCD]=x23+36−(x2−10)2
Substituting k=x2−10, the expression becomes
[A B C D]=k \sqrt{3}+\sqrt{36-k^{2}}+10 \sqrt
We can ignore the 103 for now and focus on k3+36−k2.
By the Cauchy-Schwarz inequality,
(k3+36−k2)2≤((3)2+12)((k)2+(36−k2)2)
The RHS simplifies to 122, meaning the maximum value of k3+36−k2 is 12 .
Thus the maximum possible area of ABCD is (C)12+103.
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 ∣(d−a)/3∣,∣(b−a)/3∣, and ∣(b−d)/3∣, all equal. Thus, ∣(b−a)/3∣=∣(d−a)/3∣=∣(b−d)/3∣, so ∣(b−a)∣=∣(d−a)∣=∣(b−d)∣. Hence, △DBA is equilateral.
By the Law of Cosines, [ABCD]=[ABD]+[BCD]=4(22+62−2⋅2⋅6cos(∠BCD))2⋅3+1/2⋅2⋅6sin(∠BCD).
[ABCD]=103+6(sin∠BCD−3cos(∠BCD))=103+12sin(∠BCD−60∘)≤12+103. Thus, the maximum possible area of ABCD is (C)12+103.