Spiky Problem!

Relevant wiki: Length and Area Problem Solving

Make six copies of the 1 by 8 rectangles and place them next to each as shown below, so they make a 6 by 8 rectangle. The minimum distance from the top left vertex to the bottom right vertex is a straight line, and Pythagorean theorm says that this line is 10

Let $l_1 , l_2 , \cdots , l_6$ be the length of hypotenuse of each right angles triangles (above). Clearly $l_1 = \sqrt{a^2+1}, l_2 = \sqrt{b^2+1} , \cdots, l_6 =\sqrt{f^2+1}$ To minimize the sum of the length of the 6 segments we apply AM-GM inequality $l_1+l_2+\cdots+l_6 \geq 6\sqrt[6]{l_1\cdot l_2\cdots l_5\cdot l_6}$ Equality occurs if $l_1=l_2=\cdots =l_5 =l_6 \implies a=b=\cdots = f\\ a=b=c=d=e=f=\dfrac{4}{3}\\ \implies l_1 =l_2 = \cdots =l_6 = \sqrt{\left(\frac{4}{3}\right)^2+1}=\dfrac{5}{3}$ Hence, $l_1+l_2+\cdots +l_6 \geq 6\sqrt[6]{\left(\dfrac{5}{3}\right)^6} \geq 10$ Therefore, the minimum possible answer is $\boxed{10}$ .

Minimum possible distance is when they are equally spaced out. By dividing up, we see that one length is the hypotenuse of the triangle with other sides $\frac{4}{3}$ & $1$ making one of the line segments have length $\frac53$ and the total being $6 \times \frac53 = \boxed{10}$

We see that we must move a horizontal distance of 8 and a vertical distance 6×1=6. To optimize the distance, use pythagorean's theorem: sqrt (6^2+8^2)=10.

To gain the minimum length of the sum of the six lengths, make each length the same length (unlike in the diagram where each of the lines are differmet lengths). By this I mean divide up the points where each line touches the edge of the rectangle into equal distances apart. To do this do $\frac{8}{6}$ , which is $\frac{4}{3}$ . This gives you the base length of each of these small right angled triangles formed when making each length equal. By using simple Pythagoras you can calculate the length of one of these lines:

$\sqrt{\frac{4}{3}^{2} + 1^{2}}$ = $\frac{5}{3}$

Because there are six lines all together, multiply this by six, which gives 10.

Can be solved by considering the path as the path followed by light and the sides as mirrors. Now Fermat principle (which can be easily proved) says the path is minimum for i=r. This makes all the triangles identical. and the length of each segment 5/3 and hence the total length 5/3 x 6 = 10

As the entire length has to be definitely and only be covered by 6 segments we can intuitively divide the box into 3 parts with two right-angled triangles each

• The cut off rectangle is in the dimension 2.667 * 1. We can make an equation for the sum of the two segments. √((X^2)+1) + √( $\frac{8}{3}$ -x)^2 + 1). The X is a variable and the length of the base of one of the triangles. (Can be any) Through the graph below we find the best value of x to obtain the smallest total value is 1.33. The smallest total value is 3.33. When we multiply this by 3 we get 10

Let's think about this in terms of complex numbers.
Let the sides of length $8$ and $1$ be parallel to real and imaginary axis respectively.

Now, we can think of these segments as complex numbers, $a_k \pm i$ , where $k \in [1,6]$ are real and $\sum\limits_{k=1}^{6} a_k = 8$

Now, length of the segment associated with $a_k$ is $\sqrt{a_k^2 + 1}$

However, the function $\sqrt{x^2 + 1}$ happens to be a convex function meaning that according to Jensen's inequality, $\sum\limits_{k=1}^{6} \sqrt{a_k^2 + 1} \geq 6\sqrt{\left(\dfrac16\sum\limits_{k=1}^{6} a_k\right)^2 + 1} = 10$

Equality is attained when all $a_k$ are equal meaning that the solution is $\boxed{10}$

Using Fermat's principle( link text ),we can assume that there's a beam of light going though the rectangle with total reflection.Then there will be six congruent right triangles.We can get the answer by multiplying $\frac{5}{3}$ and 6.

I disagree with the solution. Or I'm missing a criteria. If the first 5 lines are virtually vertical, their lengths are very nearly 1. The last line would then virtually cross the diagonal of the 1x8 rectangle (root 65... approx 8.06). This gives a total length approaching 13.06.

We can imagine the path of the lines as a beam of light, and we know that light always moves with the least action. We also know that the angle of incidence is equal to the angle of reflection for a light beam, so the reflected ray from every reflection is a mirrored version of the incident ray. Thus, all the lengths are identical to the first one, which traverses 8/6 units across and 1 units down, for a length of $\sqrt{1+\frac{8}{6}^2}$ . Because there are 6 identical segments, the total length is $6\sqrt{1+\frac{8}{6}^2}=\sqrt{6^2+8^2}=\sqrt{100}=10$ .

WOW for a bunch of smart people you sure make this hard.

Pythagorean Triples

A Pythagorean triple is a right triangle where all the sides are integers. One of the most popular Pythagorean triples is the 3-4-5 right triangle. 3 and 4 are the lengths of the shorter sides, and 5 is the length of the hypotenuse, the longest side opposite the right angle.

If you run through the Pythagorean Theorem on this one, you can see that it checks out:

3^1 + 4^1.33337 = 5^1.66667

3 + 5.33335 =8.33335

8.33335= 8.33335

1.66667x6=10.00002

We may think of this problem as choosing 6 distances $a_i$ such that $\sum_{i=1}^6 a_i = 8$ . The length of each segment will then be $\sqrt{a_i^2 + 1}$ and the function we want to minimize is $\sum_{i=1}^6 \sqrt{a_i^2 + 1}$ . Using lagrange multipliers, the condition for a extreme value is $\frac{a_i}{\sqrt{a_i + 1}} = \lambda$ . In other words, $\frac{a_i}{\sqrt{a_i + 1}}$ must remain invariant in our choice of distances. With this in mind let's examine the solutions of the equation

$\frac{x}{\sqrt{x^2 + 1}} = \frac{y}{\sqrt{y^2 + 1}}$ .

We may rewrite it as $x^2(y^2 + 1) = y^2(x^2 + 1) \implies (yx)^2 + x^2 = (yx)^2 + y^2 \implies x^2 = y^2 \implies |x| = |y| \implies x=y$ . The last equality only follows because we are working with distances which are positive. From this we conclude that $a_i = a_j$ . Let's say that $a_i = x$ . Then $\sum_{i=1}^6 a_i = 8 \implies 6x = 8 \implies x = \frac{4}{3}$ .

Finally, $\sum_{i=1}^6 \sqrt{a_i^2 + 1} = 6\sqrt{ \left(\frac{4}{3}\right)^2 + 1} = 10$