Problem:
Triangle ABC is a right triangle with AC=7,BC=24, and right angle at C. Point M is the midpoint of AB, and D is on the same side of line AB as C so that AD=BD=15. Given that the area of △CDM can be expressed as pmn, where m,n, and p are positive integers, m and p are relatively prime, and n is not divisible by the square of any prime, find m+n+p.
Solution:
The desired area is given by (1/2)⋅CM⋅DM⋅sinα, where α=∠CMD. The length AB=72+242=25, and, since CM is the median to the hypotenuse of △ABC,CM=25/2. Because DM is both the altitude and median to side AB in △ABD,DM=511/2 by the Pythagorean Theorem. To compute sinα, let ∠AMC=β, and note that ∠AMC and ∠CMD are complementary, so cosβ=sinα. Apply the Law of Cosines in △AMC to obtain
cosβ=2⋅225⋅225(225)2+(225)2−72=625527
The area of △CMD is (1/2)⋅CM⋅DM⋅sinα=21⋅225⋅2511⋅625527=4052711, and m+n+p=527+11+40=578.
OR
Let CH be the altitude to hypotenuse AB. Triangles CDM and HDM share side DM, and because DM∥CH,[CDM]=[HDM]=(1/2)HM⋅DM. Note that DM=AD2−AM2=2511, and that AC2=AH⋅AB. Then AH=49/25, and HM=(25/2)−(49/25)=527/50. Thus [CDM]=21⋅50527. 2511=4052711.
OR
Denote the vector CD by d and the vector CM by m. Then m=(12,7/2,0), from which d=(12−(7/25)k,7/2−(24/25)k,0), where k=(511)/2. This can be simplified to obtain d=(12−(711/10),7/2−(2411/10),0). The area of △CDM is therefore (1/2)∣m×d∣=(527/40)11.
