Converging Triangle

Say point $A_n$ has coordinates $(x_n,y_n)$ . We have

$10x_{n+3} = 3x_{n+1}+7x_n$ and $10y_{n+3} = 3y_{n+1}+7y_n$

Since we can treat $x_i$ and $y_i$ independently, we will look at the general solution of the recurrence relation $10u_{n+3} = 3u_{n+1}+7u_n$ .

The characteristic equation of this recurrence relation is $10t^3=3t+7$ . This has a real root at $t=1$ and two complex roots $t=\frac{1}{10}\left( -5 \pm 3i\sqrt{5} \right)$ (the complex roots correspond to the spiralling behaviour of the points).

In order to find point $J$ , we want to know the long-term behaviour of $u_n$ given three initial points.

Put $\tau=\frac{1}{10}\left( -5 + 3i\sqrt{5} \right)$ . The general solution of the recurrence relation is

$u_n=P+Q\tau^n+\overline{Q} \overline{\tau}^n$

(where $\overline{z}$ is the complex conjugate of $z$ . We know we can use the conjugate this way because we are dealing with real values for the coordinates)

Since $|\tau|<1$ , as $n\to \infty$ , $u_n \to P$ .

So the problem boils down to finding the value of $P$ from the initial points. Here we can save some effort by using a couple of tricks.

We could at this point solve completely generally; however, we can note that changing the initial triangle won't in fact change the ratio we are looking to find. This means we can pick a "nice" triangle and solve for that.

Further, let's relabel the vertices so that instead of starting at $n=1$ , we start at $n=0$ . This has the effect of slightly simplifying the algebra when working out $P$ .

Let's choose the starting triangle $A_0(0,0)$ , $A_1(1,0)$ , $A_2(0,1)$ . This triangle has the benefit of satisfying $y_n = x_{n-1}$ (simple to check), so $J$ lies on the line $y=x$ , and we only need to solve for $x_n$ . Since $E$ is the intersection of the lines $A_0 J$ and $A_1 A_2$ , we have $E \left( \frac{1}{2},\frac{1}{2} \right)$ .

We still have to find $J$ , however.

The three starting points give us the simultaneous equations

$P+Q+\overline{Q} = 0\\ P+Q\tau+\overline{Q} \overline{\tau} =1\\ P+Q\tau^2+\overline{Q} \overline{\tau}^2 =0$

Solving these, we find $P=\frac{10}{27}$ , and so we have $J \left(\frac{10}{27},\frac{10}{27} \right)$ .

Finally, by similar triangles (one last trick!), we simply have $\frac{A_0 J}{JE}=\frac{\frac{10}{27}}{\frac{1}{2}-\frac{10}{27}}=\frac{20}{7}$ giving the answer $20+7=\boxed{27}$ .

The given condition on the ratio of lengths, implies that

$10 A_{i+1} = 7 A_{i-2} + 3 A_{i-1} \hspace{12pt} (1)$ , which can be written as

$10 A_{i+3} - 3 A_{i+1} - 7 A_{i} = 0 , i = 1, 2, ... \hspace{24pt} (2)$

This is a difference equation of order 3, its characteristic polynomial is $10 m^3 - 3 m - 7 = 0$ , which by inspection has a root at $m = 1$ , so dividing it by $(m-1)$ using synthetic division results in $10 m^3 - 3m - 7 = (m - 1)( 10 m^2 + 10 m + 7) = 0$ . The quadratic has the roots $m = -\dfrac{1}{2} \pm i \dfrac{3 \sqrt{5}}{10}$ . It follows from this and from standard difference equation solution methods, that

$A_{k} = P + Q a^{(k-1)} \cos( \omega (k-1) ) + R a^{(k-1)} \sin ( \omega (k-1) ) \hspace{12pt} (3)$

where $a = | m | = \sqrt{0.7}$ and $\omega = \tan^{-1} \dfrac{3}{\sqrt{5}}$ , and $P, Q$ , and $R$ are two dimensional vectors. Obviously since $| a | \lt 1$ , the only vector that matters is $P$ , as the contribution of $Q$ and $P$ will die out. To solve for $P$ , we can write,

$[ A_1 , A_2, A_3 ] = [ P , Q , R ] M \hspace{12pt} (4)$

where M is the following $3 \times 3$ matrix (this is can be seen from equation (3) after substituting $k = 1, 2, 3$

$M = \begin{bmatrix} 1 && 1 && 1 \\ 1 && a \cos \omega && a^2 \cos 2 \omega \\ 0 && a \sin \omega && a^2 \sin 2 \omega \end{bmatrix} \hspace{12pt} (5)$

From (4), we now have,

$[ P, Q , R ] = [ A_1 , A_2 , A_3 ] M^{-1} \hspace{12pt} (6)$

The inverse of M can be found using any standard mathematical tool. Note that we only care about the first column of $M^{-1}$ , and this column is given by,

$M^{-1} (\cdot, 1) = \begin{bmatrix} \dfrac{7}{27} \\ \dfrac{10}{27} \\ \dfrac{10}{27} \end{bmatrix} \hspace{12pt} (7)$

So that, $P = J = \dfrac{7}{27} A_1 + \dfrac{10}{27} (A_2 + A_3) \hspace{12pt} (8)$

From this it follows that points on the line $A_1 J$ are given by $A_1 + s (A_1 J) = A_1 + s (J - A_1) = A_1 + s (- \dfrac{20}{27} A_1 +\dfrac{10}{27} (A_2 + A_3)$ )

The point $E$ depends only on $A_2$ and $A_3$ , so the value of s at point $E$ is $\dfrac{27}{20}$ , and therefore, finally, the required ratio is

$\text{Ratio} = \dfrac{1}{ \dfrac{27}{20} - 1 } = \dfrac{1}{ \dfrac{7}{20} } = \dfrac{20}{7}$

Making the answer $20 + 7 = \boxed{ 27 }$

Putting $i = 3$

In $\triangle A_1A_2A_3 \,, \; A_4 \text{ is a point on } A_1A_2 \text{ such that } \Large\frac{A_1A_4}{A_4A_2} = \frac{3}{7}$

$\text{This pattern is continued for } i\geq 4 \text{ and we will get a point } J \text{ which is the converging point of } \triangle A_{i-2}A_{i-1}A_i \text{ for } i \rightarrow \infty .$

$\text{Join } A_3 \text{ and } J \text{ and extend it to meet } A_1A_2 \text{ at } F_3.$

$\text{Now taking one extra case of }i = 2 \text{ and assuming some point }A_0 \text{ in the plane of figure. } \text{ According to question, we will have a line }A_2A_3 \text{ with } A_3\text{ lying on line } A_0A_1 \text{ such that }$

$\hspace{8cm}\Large\frac{A_0A_3}{A_3A_1} = \frac{3}{7}$ $\qquad\cdots \;Eq. 1$

That means $A_0 \text{ lies on } A_1A_3 \text{ extended and located such that } Eq. 1 \text{ holds. Join }A_2 \text{ and }A_0.\text{ Join }A_2\text{ and }J \text{ and extend it to meet }A_1A_3\text{ at }F_2.\text{ This is illustrated in figure below. }$

$\text{We see that adding one extra point } A_0 \text{ or in fact, forming a new }\triangle A_0A_1A_2 \text{ does not affect the position of }J. \text{ So we can conclude that } \triangle A_0A_1A_2 \text{ is analogous to }\triangle A_1A_2A_3 \text{ if we take the starting triangle as } \triangle A_0A_1A_2. \text{ So, line } A_2JF_2\text{ is analogous to line }A_3JF_3.$

$\text{So, }\Large\frac{A_2J}{JF_2} =\frac{A_3J}{JF_3}$ $\qquad\cdots \;Eq. 2$

$\text{It can be easily prove that because of }Eq. 2\,,\; J\text{ lies on median of } \triangle A_1A_2A_3 \text{ through }A_1. \text{ I am skipping this proof. }$

$\text{Join }A_1 \text{ and } J \text{ and extend it to meet }A_2A_3 \text{ at } E.\text{ Then }E\text{ is the mid-point of }A_2A_3.$

$\text{In the similar manner taking }i=4 \text{ we get }\triangle A_2A_3A_4 \text{ analogous to }\triangle A_1A_2A_3 \text{ and also line } A_4JF_4 \text{ analogous to line } A_3JF_3.$

$\text{So, }\Large\frac{A_4J}{JF_4} =\frac{A_3J}{JF_3}$ $\qquad\cdots \;Eq. 3$

$\text{Again, because of }Eq. 3\,,\; J\text{ lies on median of } \triangle A_2A_3A_4 \text{ through }A_2.\text{ which coincides with the line }A_2JF_2.$

$\text{Let this line intersect }A_3A_4 \text{ at }H.\text{ Then }H\text{ is the mid-point of }A_3A_4.$

$\text{Thus, location of }J\text{ can be defined as the intersection point of two lines }A_1E \text{ and }A_2H\text{ where } E \text{ and }H \text{ are the mid-point of }A_2A_3 \text{ and }A_3A_4 \text{ respectively. This is shown in figure below.}$

$\text{Let area of }\triangle A_1A_2A_3 \text{ be } X. \text{ Further in the solution below, area of any triangle }PQR \text{ will be denoted by }ar(\triangle PQR).$

$\text{I have frequently used the property of the triangle that the area of triangles between the same parallels are in the proportions of their base. }$

$\Large\frac{ar(\triangle A_3A_1A_2)}{ar(\triangle A_3A_1A_4)} = \frac{A_1A_2}{A_1A_4} = \frac{A_1A_4+A_4A_2}{A_1A_4} = 1+\frac{7}{3}=\frac{10}{3}$

$\Rightarrow ar(\triangle A_3A_1A_4) = \frac{3X}{10}$

$\Rightarrow ar(\triangle A_3A_4A_2) = X-\frac{3X}{10} = \frac{7X}{10}$

$\Large\Rightarrow \frac{ar(\triangle A_3HA_2)}{ar(\triangle A_3A_4A_2)} = \frac{1}{2}$

$\Rightarrow ar(\triangle A_3HA_2) = \frac{1}{2}\cdot ar(\triangle A_3A_4A_2) = \frac{1}{2}\cdot \frac{7X}{10} = \frac{7X}{20} \hspace{1cm}\cdots \;Eq. 4$

$\Rightarrow ar(\triangle A_1HA_2) = ar(\triangle A_1EA_2) = \frac{1}{2}\cdot ar(\triangle A_3A_1A_2)=\frac{X}{2}\hspace{1cm}\cdots\; Eq. 5 \qquad [\because H\,,\,E \text{ are the mid-points of }A_3A_4\text{ and }A_3A_2 \Rightarrow HE \parallel A_1A_2]$

$\text{Now, }$ $\Large\frac{A_3F_2}{F_2A_1}=\frac{ar(\triangle A_3F_2A_2)}{ar(\triangle A_1F_2A_2)}=\frac{ar(\triangle A_3F_2H)}{ar(\triangle A_1F_2H)}=\frac{ar(\triangle A_3F_2A_2)-ar(\triangle A_3F_2H)}{ar(\triangle A_1F_2A_2)-ar(\triangle A_1F_2H)} = \frac{ar(\triangle A_3HA_2)}{ar(\triangle A_1HA_2)}=\frac{\frac{7X}{20}}{\frac{X}{2}}=\frac{7}{10}$ $\hspace{1cm}\cdots\;Eq. 6$ $\text{ }[\text{ From }Eq. 4\text{ and }Eq.5 ]$

$\text{Also, }$ $\Large\frac{A_3F_2}{F_2A_1}=\frac{ar(\triangle A_3F_2A_2)}{ar(\triangle A_1F_2A_2)}=\frac{ar(\triangle A_3F_2J)}{ar(\triangle A_1F_2J)}=\frac{ar(\triangle A_3F_2A_2)-ar(\triangle A_3F_2J)}{ar(\triangle A_1F_2A_2)-ar(\triangle A_1F_2J)}=\frac{ar(\triangle A_3JA_2)}{ar(\triangle A_1JA_2)}$

$\Large\Rightarrow \frac{ar(\triangle A_3JA_2)}{ar(\triangle A_1JA_2)}=\frac{A_3F_2}{F_2A_1}=\frac{7}{10}$ $\hspace{1cm}[\text{From }Eq. 6]$

$\Large\Rightarrow \frac{2\cdot ar(\triangle EJA_2)}{ar(\triangle A_1JA_2)}=\frac{7}{10}$ $\hspace{1cm}\Bigg[\because\frac{ar(\triangle A_3JA_2)}{ar(\triangle EJA_2)}=2\Bigg]$

$\Large\Rightarrow \frac{ar(\triangle EJA_2)}{ar(\triangle A_1JA_2)}=\frac{7}{20}$

$\text{Now, }$ $\Large\frac{A_1J}{JE}=\frac{ar(\triangle A_1JA_2)}{ar(\triangle EJA_2)}=\frac{20}{7}$

$\text{So, the answer is }20+7=27$