Problem:
Let △ABC be a right triangle in the xy-plane with the right angle at C. Given that the length of the hypotenuse AB is 60, and that the medians through A and B lie along the lines y=x+3 and y=2x+4, respectively, find the area of △ABC.
Solution:
More generally, we will show that if △ABC is a right triangle with right angle at C, if AB=2r and if the acute angle between the medians emanating from A and B is θ, then
area(△ABC)=34r2tanθ(1)
In our case, r=260=30 and tanθ=31 (determined either by the standard formula for the tangent of the angle between two given lines, or by glancing at the first accompanying figure), so the answer to the problem is (34)(30)2(31) or 400. To establish (1), first note that
area(△ABC)=r2sinψ,(2)
where ψ=∠AOC, as shown in the second figure below, where 0 is the midpoint of AB,D is the centroid of △ABC and CO=r (since △ABC is a right triangle). Consequently, to prove (1), it suffices to verify that sinψ=(34)tanθ, which is equivalent to establishing that
cosϕ=−(34)tanθ,(3)
where ϕ=ψ+90∘. This consideration leads us to the third figure below, where O′ is the center of the circle through D and B, and ∠DOO′=ϕ.
To prove (3), we will apply the Law of Cosines to △DOO′. (The fact that ∠OO′B=θ comes from the observation that ∠ADB=180∘−θ, and hence arc ADB has central angle 2θ; DO=3r is a well-known fact concerning the centroid.) Noting that DO′=BO′=rcscθ and OO′=rcotθ (from △BOO′), indeed,
As in the diagram shown on the right, Let M,N and O be the midpoints of the sides of △ABC,D be its centroid, p,q,s and t be the distances indicated, and θ=∠ADM. As in the previous solution, also observe that tanθ=31, and hence sinθ=101. Then from area (△ABC)=6⋅area(△ADM),
area(△ABC)=6st sinθ=106st, (1)
and, since △ABC is a right triangle,
area(△ABC)=2pq(2)
Moreover, by the Pythagorean Theorem, we find that
4p2+4q2p2+4q24p2+q2=AB2=3600=9s2=9t2
With the help of these expressions from (1) and (2) it follows that
