Length of race track based on the final distance between participants

Given three integers A, B, and C, the task is to find the length of a race track if 3 racers are competing in a race where the first racer beats the second racer by A meters, the first racer beats the third racer by B meters and the second racer beats the third by C meters.

Examples:

Input: A = 11, B = 90, C = 80
Output: 880
Input: A = 10, B = 20, C = 12
Output: 60

Approach :
Let X be the length of the race track.

Case 1: By the time when the First racer finished the race, the distances covered by all the 3 racers are:
First = X, Second = X – A, Third = X – B
Let the time taken by the First racer to finish the race be T₁.

Case 2: By the time when the Second racer finished the race, the distances covered by the remaining 2 racers are:
Second = X, Third = X – C
Let the time taken by the Second racer to finish the race be T₂.

The ratio of the speeds of the Second and the Third racer will be constant in both case 1 and case 2 which implies:

=> ((X – A) / T₁) / ((X – B) / T₁) = (X / T₂) / ((X – C) / T₂)
=> (X – A) / (X – B) = (X) / (X – C)
=> X² – A*X – C*X + A*C = X² – B*X
=> A*C = (C + A – B)*X
=> X = A*C / (C + A – B)

Below is the implementation of the above program:

C++

// C++ Program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
#define int long long
 
int32_t main()
{
    int A = 11;
    int B = 90;
    int C = 80;
 
    int ans = C * A;
    ans = ans / (C + A - B);
 
    cout << ans << endl;
}

Java

// Java Program for the
// above approach
 
import java.util.Scanner;
 
class GFG {
    public static void main(String args[])
    {
        int a = 11;
        int b = 90;
        int c = 80;
 
        System.out.println(c * a
                           / (c + a - b));
    }
}

Python3

# Python3 Program for the
# above approach
 
# Function to get the length
# of the race track
def findlength(a, b, c):
    # return the answer
    return c * a/(c + a-b)
 
a = 11
b = 90
c = 80
 
print(findlength(a, b, c))

C#

// C# program for the above approach
using System;
class GFG{
     
static void Main()
{
    int a = 11;
    int b = 90;
    int c = 80;
         
    Console.WriteLine(c * a / (c + a - b));
}
}
 
// This code is contributed by divyeshrabadiya07

Javascript

<script>
 
// JavaScript program for the above approach
   
// Driver Code
let a = 11;
let b = 90;
let c = 80;
 
document.write(c * a / (c + a - b));
 
// This code is contributed by sanjoy_62   
 
</script>