CC000 - Editorial

editorial
, ,
1

Contest Name:

CODICTION 2020

PROBLEM LINK:

Problem
Contest

Author: Srijan Agrawal
Tester: Mradul Rathore, Nehal Jain, Saniya Agrawal
Editorialist: Murtaza Ali

DIFFICULTY:

CAKEWALK.

PREREQUISITES:

Geometry.

PROBLEM:

Find a point on x-axis such that the reflection of (x_{2}, y_{2}) can be seen from the point (x_{1}, y_{1}).

QUICK EXPLANATION:

We have to place a mirror on (x, 0) such that the reflection of (x_{2}, y_{2}) in the mirror on (x, 0) can be seen from the point (x_{1}, y_{1}).

EXPLANATION:

The Law of Reflection states that the angle of incidence is equal to the angle of reflection. Let the ray passing through (x_{2}, y_{2}) and (x, 0) be the incident line and the ray from (x_{1}, y_{1}) and (x, 0) be the reflected line.

Let,
Angle of incidence = Angle of reflection = \theta

Slope of line of incidence = tan(\frac{\pi}{2} - \theta) = cot(\theta)

Slope of line of reflection = tan(\frac{\pi}{2} + \theta) = - cot(\theta)

Slope of line of incidence in two point form = \frac{y_{2} - 0}{x_{2} - x} = \frac{y_{2}}{x_{2} - x}

Slope of line of reflection in two point form = \frac{y_{1} - 0}{x_{1} - x} = \frac{y_{1}}{x_{1} - x}

By above equations, we can get,
\frac{y_{2}}{x_{2} - x} = - \frac{y_{1}}{x_{1} - x}

x = \frac{x_{2}*y_{1} + x_{1}*y_{2}}{y_{1} + y_{2}}

COMPLEXITY:

Time complexity: O(1) per test.

SOLUTIONS:

Setter's Solution 
#include<bits/stdc++.h>
using namespace std;

int main()
{
    int T;
    cin>>T;
    while(T--){
        long long int x1,y1,x2,y2;
        cin>>x1>>y1>>x2>>y2;
        double n=(x2*y1+y2*x1);
        double ans=n/(y1+y2);
        cout << fixed << setprecision(2)<<ans<<endl;
    }
    return 0;
}

ALTERNATE SOLUTION:

We can also solve this problem using Similarity of Triangles. If you want detailed explanation for this approach please let us know.

Alternate Solution by Tester 
#include <iostream>
using namespace std;

int main() {
int t;
cin>>t;
while(t--){
    int x1, y1, x2, y2;
    cin>>x1>>y1;
    cin>>x2>>y2;
    float r = (float) abs(x1 - x2) / (y1 + y2);
    if(x1 < x2){
        r *= y1;
        r += x1;
    }
    else{
        r *= y2;
        r += x2;
    }
    printf("%0.2f\n", r);
}
return 0;
}

Feel free to share your approach. If you have any queries, they are always welcome.

1 Like