情趣生活
情趣生活 => 灵机一动 => Topic started by: MrPuzzle on 七月 22, 2004, 12:58:11 pm

Here's another geometric problem that got my attention (i.e. interesting).

Triangle ABC has sides of length 13,14,15. Triangle PQR is inside triangle ABC such that the corresponding sides are parallel, and two units away. Find the area of triangle PQR.


It is quite an interesting problem. Like the twoladder one, seems simple, yet require brutal force to solve. And there's a trick here before set up the formulas.
Since the distance of each line of PQR is the same to ABC (2 units), then points P, Q, R are on the 角平分线 of angles A, B, C. That is the key to solve the problem. The rest is just to use brutal force to calculate. .......

Different from the 2 ladder problem, this one doesn't need heavy computer computation. If you set up everything correctly, it can be solved in 3 or 4 lines. I assume everybody knows the area formula:
A = sqrt(s*(sa)*(sb)*(sc)) where s = (a+b+c)/2 and a,b,c are the side length.

hmm.. even simpler?
Notice that big triangle ABC ~ small triangle (denoted as) abc.
Denote the ratio of perimeter of ABC (P) and that of abc (p) as r.
Then area of abc (s) = r*r*S, S is the area of ABC.
However, abc + 3 trapz.'s around it (connecting the vertex of the triangles), you will get ABC. The area of the 3 trapz. is easy to find by:
1/2(A+a)*2+1/2(B+b)*2+1/2(C+c)*2 (here just use the same letter for the sides..) = P+p.
Hence S=P+p+s
But P=13+14+15=42, p=r*P=42r, S = 84, s= r*r*S
84*r*r+42*r+42=84
r=1/2
s=21

这个题目最近又出现了。我想起以前在这个论坛见过这个题目，找了一下，果然找到了。当时忘了贴我的答案，现在贴出：
由Heron公式，很容易算出ABC的面积是84. ABC的中心（角平分线的焦点）到各边等距，设此距离是H。ABC的面积等于H乘边长和的一半，所以H等于4. 把各边向内推进2得到小三角形。2是4的一半。所以小三角形面积是ABC面积的1/4，s = 21.