LOJ-1056: Olympics

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What the problem wants : The problem wants you to find the height and width of a field which will have a ratio of a : b and perimeter of 400 meters. The field is a rectangle with two sliced circle at two opposing ends.

Analyzing the shape : Due to the unique shape of the perimeter of it will be 2*(L + S), where L = Length of the field and S = Width of the field (take note that here width isn't the width of the rectangle rather its the circumference S. Since the perimeter of a circumference is r*θ. Look at the below figure to get a better idea.

As we can see from the above figure , the field is ABCD . Where, AD = BC = L and AB = CD = W (for the rectangle) . Also AB = CD = S (For Circumference ) and the center of the circle is point "O".

General Approach to the solution : Since we have a fixed value as perimeter we can use it in an equation for deriving the width and length.

The equation will be, 2*(L + S) = 400

Lets look at the above figure of the field for deriving more information. Here , OA = OB = OC = OD, since all four are the radius for the circle.

As we can see that radius, r = length² + width²

the angle for circumference of length "S" is, θ = cos^-1((2.0 *radius²)/(2.0 * radius²))

From previous equation 2*(L + S) = 400 , we can get 2*(L + r*θ) = 400

From triangle ADC we get, (2r)² = W² + L² (Using Pythagoras' theorem)

or , r = √(W² + L²)/2

From S = r * θ we can get cosθ = 2r² - W² * 2r² (Using law of cosines)

Let , L/W = x and the provided ration of L : W = a : b.

Hence, 2(L + S) = 400

or, 2(ax + xrθ) = 400

or , x = 400/(2a + 2S)

Now the answer will be x into Length = a * x , Width = b * x.

Resources :

1. Circumference
2. Pythagoras' theorem
3. Law of cosines

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Solution Code in cpp

In the below code, acos( ) function returns the cos^-1( ) of the passed parameter . Variables were taken and calculated in double format to ensure high precision calculation.

#include<bits/stdc++.h>
using namespace std;

int main()
{

    double n,a,b;
    int t,cas=1;
    cin>>t;

    while(t--)
    {
        scanf("%lf : %lf",&a,&b);

        //pythagorous theorem
        double radius=sqrt(a*a+b*b)/2.0;

        //law of cosine
        double theta=acos((2.0*radius*radius-b*b)/(2.0*radius*radius));

        // circumference
        double s=radius*theta;

        double x=400.0/(2*a+2*s);

        //as given in question
        double length=x*a,width=x*b;

        printf("Case %d: %.8f %.8f\n",cas++,length,width); // "%.8f" formats the output for printing 8 digits after decimal point

    }
}

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Tutorial source