# 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

Title: Area of a triangle
Post by: MrPuzzle on 七月 22, 2004, 12:58:11 pm
Here's another geometric problem that got my attention (i.e. interesting).

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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.
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Title: Area of a triangle
Post by: packman on 七月 23, 2004, 09:11:47 am
It is quite an interesting problem. Like the two-ladder 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. .......
Title: Area of a triangle
Post by: MrPuzzle on 七月 23, 2004, 09:56:43 am
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*(s-a)*(s-b)*(s-c))  where s = (a+b+c)/2 and a,b,c are the side length.
Title: Area of a triangle
Post by: idiot94 on 七月 23, 2004, 12:20:42 pm
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
Title: Re: Area of a triangle
Post by: 万精油 on 一月 19, 2018, 04:46:39 pm