can someone help me out in this question

Let’s call the answer l. Now, If you see, as l increases, h decreases. Let’s calculate the value of h for some given l.

We start with writing the lines in co-ordinate form, let us assume the start of the bottom line is (0,0).

The first line passes through (0,0) and (l, \sqrt{a^2 - l^2}) by pythagoras

theorem.The second line passes through the points (0, \sqrt{b^2 - l^2}) and (l,0).

We can now find their equations.

Line 1 : y = x\sqrt{a^2 - l^2}/l

Line 2 : y = (l-x)\sqrt{b^2 - l ^2}/l

Removing x using the 2 equations, gives us

y = \sqrt{a^2 - l^2}\sqrt{b^2 - l^2}/(\sqrt{a^2 - l^2} + \sqrt{b^2 - l^2}).

The y co-ordinate of their intersection is h.

Now we binary search over l. If y is too small, then we decrease l, else we increase it.

One important thing to note, is that if h>ab/(a+b), then there is no solution because that’s the answer for l=0, and since l can’t be negative, and the answer is largest at 0, so h would be larger than the largest possible value.

## My code

```
#include <iostream>
#include <bits/stdc++.h>
#include <cmath>
#include <vector>
#define ll long long int
#define mp make_pair
#define pb push_back
#define vi vector<int>
using namespace std;
long double a,b,h;
const long double eps=1e-5;
long double calc(long double l){
long double x=sqrtl(b*b - l*l);
long double y=sqrtl(a*a - l*l);
return (x*y)/(x+y);
}
void solve(){
cin>>a>>b>>h;
long double mn=0;
long double mx=min(a,b);
long double l=(mn + mx)/2;
if(h > (a*b)/(a+b)){
cout<<"-1\n";
return;
}
while(mx-mn > eps/10){
if(calc(l)>h){
mn=l;
l = (l+mx)/2;
}
else{
mx=l;
l=(l+mn)/2;
}
}
cout<<fixed<<setprecision(9)<<(mn+mx)/2<<"\n";
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin >>t;
while(t--){
solve();
}
}
```