Fibonacci Triangles

First, since both $PS$ and $TS$ are perpendicular to the perpendicular bisector of $QR$ , then any point $T$ satisfies the given condition must be on the line $PS$ . Furthermore, $PS$ is parallel to $QR$ , so $d(P,QR)=d(T,QR)$ . This implies that the operation conserve the area of the triangle $PQR$ .

Now, we need to find the area of the original triangle $PQR$ . Assuming that $P(F_{30},F_{31}); Q(F_{32},F_{33}); R(F_{34},F_{35})$ , we have: $\overrightarrow{QP}=(F_{30}-F_{32};F_{31}-F_{33})=(-F_{31},-F_{32})$ $\overrightarrow{QR}=(F_{34}-F_{32};F_{35}-F_{33})=(F_{33},F_{34})$

Then $[PQR]=\frac{1}{2}|\overrightarrow{QP}*\overrightarrow{QR}|=\frac{1}{2}|-F_{31}F_{34}+F_{32}F_{33}|$ $=\frac{1}{2}|F_{32}(F_{31}+F_{32})-F_{31}(F_{32}+F_{33})|$ $=\frac{1}{2}|F_{32}(F_{31}+F_{32})-F_{31}(F_{32}+F_{33})|$ $=\frac{1}{2}|F_{32}^2-F_{31}F_{33}|=\frac{1}{2}|F_{30}F_{32}-F_{31}^2|=...$ $=\frac{1}{2}|F_{2}^2-F_{1}F_{3}|=\frac{1}{2}|1*1-1*2|=\frac{1}{2}$ (we can use Heron's formula for the same result)

Next, put $P'(x,y); Q'(0,0); R'(100,0)$ where $x,y \leq 0$ (since $P'$ is in the first quadrant). The two points are on the line $y=0$ . Since $P'Q'=100$ , then the set of point P' satisfies the condition $[P'Q'R']=[PQR]$ is the line $\Delta$ such that $\Delta$ is parallel to $Ox$ and $d(\Delta,Ox)=\frac{2[P'Q'R']}{PQ}=\frac{1}{100}$ . Hence, we have 2 possibilities of $\Delta$ : $y=\pm\frac{1}{100}$ . Then the largest possible value of $y$ is $\frac{1}{100}$ , which yields the answer 101.

Let M be the midpoint of QR. Since S is the shortest point from P to the perpendicular bisector of QR, we have PS perpendicular to SM $\Rightarrow$ PS parallel to QR. Similarly, TS is also parallel to QR. Hence P, S and T are collinear and PST is parallel to QR $\Rightarrow$ PQR and TQR have the same base (QR) and height $Rightarrow$ PQR and TQR have the same area.

The invariant in this transformation is the area of the triangle. $F_{30}$ = 832040, $F_{31}$ = 1346269, $F_{32}$ = 2178309, $F_{33}$ = 3524578, $F_{34}$ = 5702887, $F_{35}$ = 9227465. Hence the area of the triangle = $\frac{1}{2}|F_{30} \cdot F_{33} + F_{32} \cdot F_{35} + F_{34} \cdot F_{31} - F_{32} \cdot F_{31} - F_{34} \cdot F_{33} - F_{30} \cdot F_{35}| = \frac{1}{2}$ .

Since the base (x-axis) in the question is 100, and the area is $\frac{1}{2}$ , we have the height (y-axis) = $\frac{\frac{1}{2}}{100} \cdot 2 = \frac{1}{100}$ , hence the answer is 101.

S=1/2|(1,F30,F31),(1,F32,F33),(1,F34,F35)|=1/2|(0,F31,F32),(1,F32,F33),(0,F33,F34)|=1/2|F31 F34-F32 F33|=1/2

Notice that the condition implies that P must remain at the same height from QR, so by the half base times height formula, the area of PQR is constant.

We can work out what it is at the start, given the determinant formula for the area in Cartesian coordinates, by writing the Fibonacci numbers in terms of $\phi$ , and explicitly computing the determinant in terms of $\phi$ . In fact, the answer comes out to 1.

Hence, the area of the triangle is $\frac{1}{2}$ , meaning that the y-coordinate must be $\frac{1}{100}$ , giving the answer of 101.

The answer is the same, regardless of which Fibonacci number you start out with. In the following, we will replace $30$ with $k$ .

The description of what move we are allowed to do can be simplified. We are allowed to choose any point $P$ and move it along the line through $P$ that is parallel to the line between the other pair of points. Notice that this does not change the area of the triangle, since we can think of the base being the side between the pair of unchanging points, and the height to the third point does not change when it moves in the prescribed manner.

To determine the $y$ -coordinate of the third point in the final position we need to determine the area of the triangle.

We can use the coordinate geometry formula for area of a triangle. $A = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1\\ \end{vmatrix}$

Note that this is equivalent to the formula $A = \left|\frac{x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)}{2} \right|$ , which we will use for simplicity.

We can reduce our number of variables by defining $F_{k+2}, F_{k+3}, F_{k+4}, F_{k+5}$ in terms of $a = F_{k}$ and $b = F_{k+1}$ . Our area expression will become

$A = \left |\frac{a((a+2b) - (3a+5b)) + (a+b)((3a+5b) - b) + (2a+3b)(b - (a+2b))}{2} \right| = \left |\frac{b^2 - a^2 - ab}{2}\right | = \left|\frac{b^2 - (a+b)(a)}{2} \right|$

Substituting our Fibonacci numbers back in, we have $A = \left|\frac{F_{k+1}^2 -F_k F_{k+2}}{2}\right|$ . The top expression is a well-known Fibonacci identity (Cassini's identity), with value equal to $\pm 1$ . So the area of the triangle is always $\frac{1}{2}$ , independent of the value of $k$ .

Let $h$ be the $y$ -coordinate of the third point from the question. Then $\frac{100h}{2} = \frac{1}{h}$ so $h = \frac{1}{100}$ and $a + b = 1 + 100 = 101$ .