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 β³CDM can be expressed as pmnββ, 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)β CMβ DMβ sinΞ±, where Ξ±=β CMD. The length AB=72+242β=25, and, since CM is the median to the hypotenuse of β³ABC,CM=25/2. Because DM is both the altitude and median to side AB in β³ABD,DM=511β/2 by the Pythagorean Theorem. To compute sinΞ±, let β AMC=Ξ², and note that β AMC and β CMD are complementary, so cosΞ²=sinΞ±. Apply the Law of Cosines in β³AMC to obtain
The area of β³CMD is (1/2)β CMβ DMβ sinΞ±=21ββ 225ββ 2511βββ 625527β=4052711ββ, and m+n+p=527+11+40=578β.
OR
Let CH be the altitude to hypotenuse AB. Triangles CDM and HDM share side DM, and because DMβ₯CH,[CDM]=[HDM]=(1/2)HMβ DM. Note that DM=AD2βAM2β=2511ββ, and that AC2=AHβ AB. Then AH=49/25, and HM=(25/2)β(49/25)=527/50. Thus [CDM]=21ββ 50527β. 2511ββ=40527β11β.
OR
Denote the vector CD by d and the vector CM by m. Then m=(12,7/2,0), from which d=(12β(7/25)k,7/2β(24/25)k,0), where k=(511β)/2. This can be simplified to obtain d=(12β(711β/10),7/2β(2411β/10),0). The area of β³CDM is therefore (1/2)β£mΓdβ£=(527/40)11β.
