CEMC Fermat

In this solution, we will use two geometric results: (i) The Triangle Inequality This result says that, in $\triangle A B C$, each of the following inequalities is true: $A B+B C>A C \quad A C+B C>A B \quad A B+A C>B C$. This result comes from the fact that the shortest distance between two points is the length of the straight line segment joining those two points. For example, the shortest distance between the points $A$ and $C$ is the length of the line segment $A C$. Thus, the path from $A$ to $C$ through a point $B$ not on $A C$, which has length $A B+B C$, is longer. This explanation tells us that $A B+B C>A C$. (ii) The Angle Bisector Theorem In the given triangle, we are told that $\angle Q R T=\angle S R T$. This tells us that $R T$ is an angle bisector of $\angle Q R S$. The Angle Bisector Theorem says that, since $R T$ is the angle bisector of $\angle Q R S$, then $\frac{Q T}{T S}=\frac{R Q}{R S}$. The Angle Bisector Theorem can be proven using the sine law: In $\triangle R Q T$, we have $\frac{R Q}{\sin (\angle R T Q)}=\frac{Q T}{\sin (\angle Q R T)}$. In $\triangle R S T$, we have $\frac{R S}{\sin (\angle R T S)}=\frac{T S}{\sin (\angle S R T)}$. Dividing the first equation by the second, we obtain $\frac{R Q \sin (\angle R T S)}{R S \sin (\angle R T Q)}=\frac{Q T \sin (\angle S R T)}{T S \sin (\angle Q R T)}$. Since $\angle Q R T=\angle S R T$, then $\sin (\angle Q R T)=\sin (\angle S R T)$. Since $\angle R T Q=180^{\circ}-\angle R T S$, then $\sin (\angle R T Q)=\sin (\angle R T S)$. Combining these three equalities, we obtain $\frac{R Q}{R S}=\frac{Q T}{T S}$, as required. We now begin our solution to the problem. By the Angle Bisector Theorem, $\frac{R Q}{R S}=\frac{Q T}{T S}=\frac{m}{n}$. Therefore, we can set $R Q=k m$ and $R S=k n$ for some real number $k>0$. By the Triangle Inequality, $R Q+R S>Q S$. This is equivalent to the inequality $k m+k n>m+n$ or $k(m+n)>m+n$. Since $m+n>0$, this is equivalent to $k>1$. Using the Triangle Inequality a second time, we know that $R Q+Q S>R S$. This is equivalent to $k m+m+n>k n$, which gives $k(n-m)<n+m$. Since $n>m$, then $n-m>0$ and so we obtain $k<\frac{n+m}{n-m}$. (Since we already know that $R S>R Q$, a third application of the Triangle Inequality will not give any further information. Can you see why?) The perimeter, $p$, of $\triangle Q R S$ is $R Q+R S+Q S=k m+k n+m+n=(k+1)(m+n)$. Since $k>1$, then $p>2(m+n)$. Since $2(m+n)$ is an integer, then the smallest possible integer value of $p$ is $2 m+2 n+1$. Since $k<\frac{n+m}{n-m}$, then $p<\left(\frac{n+m}{n-m}+1\right)(n+m)$. Since $n+m$ is a multiple of $n-m$, then $\left(\frac{n+m}{n-m}+1\right)(n+m)$ is an integer, and so the largest possible integer value of $p$ is $\left(\frac{n+m}{n-m}+1\right)(n+m)-1$. Every possible value of $p$ between $2 m+2 n+1$ and $\left(\frac{n+m}{n-m}+1\right)(n+m)-1$, inclusive, can actually be achieved. We can see this by starting with point $R$ almost at point $T$ and then continuously pulling $R$ away from $Q S$ while keeping the ratio $\frac{R Q}{R S}$ fixed until the triangle is almost flat with $R S$ along $R Q$ and $Q S$. We know that the smallest possible integer value of $p$ is $2 m+2 n+1$ and the largest possible integer value of $p$ is $\left(\frac{n+m}{n-m}+1\right)(n+m)-1$. The number of integers in this range is $\left(\left(\frac{n+m}{n-m}+1\right)(n+m)-1\right)-(2 m+2 n+1)+1$. From the given information, the number of possible integer values of $p$ is $m^{2}+2 m-1$. Therefore, we obtain the following equivalent equations: $\left(\left(\frac{n+m}{n-m}+1\right)(n+m)-1\right)-(2 m+2 n+1)+1 =m^{2}+2 m-1$. Solving this, we find $n-m=4$.