First, by Pythagoras,. Next, we solve for by computing the area of triangle in XD AMD two different ways. Using as the base, we find the area to be . Using as the base, AD AM we find the area to be . Thus, and hence . Next, applying Pythagoras to triangle we get. Finally, note that since = XMD, ∠DXM = 90◦, the quadrilateral is cyclic. Therefore, we may compute the product DCMX ∠DCM via Ptolemy’s theorem: and . XC · MD XC · MD = XM · DC + XD · MC Let . Then, applying the Law of Cosines, we get . Then apply the Law of Cosines to triangle and we get ADC Option 1: Consider a transformation that scales the major axis by a factor of ½ so that the ellipse becomes a circle of radius 1 and the equilateral triangle becomes an isosceles triangle. Let the transformed triangle be denoted . Now, let . Then the original length |BC| = 4x and since is equilateral, |AB| = |BC| = 4x. Next, we calculate the altitude dropped from as. Applying Pythagoras’s theorem, we get that . Substitute |AB| = 4x in and solve for . We get that . Finally, we calculate the area of. Since is equilateral, its area is given by . Substituting in for , we get . Option 2: Write the equation of the ellipse: . Let the side of the equilateral triangle have length 2m. Then a vertex of the triangle would fall on . Put this ordered pair into the equation of the ellipse and you get that . Then the area is For fun: I got: 91411 + 1 + 2 or 91411 + 3
First, by Pythagoras,. Next, we solve for XD by computing the area of triangle AMD in two different ways. Using AD as the base, we find the area to be . Using AM as the base, we find the area to be . Thus, and hence . Next, applying Pythagoras to triangle XMD, we get. Finally, note that since ∠DXM = ∠DCM = 90◦, the quadrilateral DCMX is cyclic. Therefore, we may compute the product XC · MD via Ptolemy’s theorem: XC · MD = XM · DC + XD · MC and . Let . Then, applying the Law of Cosines, we get . Then apply the Law of Cosines to triangle ADC and we get Option 1: Consider a transformation that scales the major axis by a factor of ½ so that the ellipse becomes a circle of radius 1 and the equilateral triangle becomes an isosceles triangle. Let the transformed triangle be denoted . Now, let . Then the original length |BC| = 4x and since is equilateral, |AB| = |BC| = 4x. Next, we calculate the altitude dropped from as. Applying Pythagoras’s theorem, we get that . Substitute |AB| = 4x in and solve for . We get that . Finally, we calculate the area of. Since is equilateral, its area is given by . Substituting in for , we get . Option 2: Write the equation of the ellipse: . Let the side of the equilateral triangle have length 2m. Then a vertex of the triangle would fall on . Put this ordered pair into the equation of the ellipse and you get that . Then the area is For fun: I got: 91411 + 1 + 2 or 91411 + 3