Inscribed perimeter

It is a standard result that the inscribed triangle in an acute-angled triangle with the shortest perimeter is the orthic triangle, which has the feet of the altitudes as vertices.

This triangle is acute-angled (just), and the orthic triangle has side lengths $a\cos A$ , $b\cos B$ , $c\cos C$ . Thus the minimum perimeter is $P \; = \; a \frac{b^2 + c^2 - a^2}{2bc} + b\frac{a^2 + c^2 - b^2}{2ac} + c\frac{a^2 + b^2 - c^2}{2ab} \; = \; \frac{2(a^2b^2 + a^2c^2 + b^2c^2) - (a^4 + b^4 + c^4)}{2abc}$ which makes the minimum perimeter $\boxed{8.96}$ for this triangle.

We can fix the vertices on the 9 and 10 sides as shown in the picture and then determine where to put the third vertex on the 5 side such that the sum of the two new sides is minimized. The sum of the two new sides is minimized when the angles with the 5 side are equal. This result is made obvious by reflecting one of the points across the 5 side (as shown in the second picture), and seeing that the shortest distance between any two points is a straight line.

We can label the other two pairs of angles in the same way. Notice that the three straight angles have to add up to 540 degrees. Subtracting off the angles of the inscribed triangle and dividing by 2 tells us that the three angles we marked must add up to 180 degrees. With this fact we can label the angles of the of the original triangle. We have four similar triangles, so we can label the scaling factors of the three new triangles $i, j,$ and $k$ (scaled from the original triangle). By similar triangles, we can solve for the sides of the original triangle to get $\large 5i+10k=9, 9k+5j=10, 10j+9i=5$ . Solving this system gives $\large i=\frac{1}{15}, j=\frac{11}{25}, k=\frac{13}{15}$ . The perimeter of the inscribed triangle is $\large 10i+9j+5k$ , so the answer is $\large \boxed{8.96}$

See Henning's explanation about the orthic triangle and his following comment below. In this figure, the blue line $D_1 D_2$ is the straight path through the refections of the triangle, so that it makes equal angles with the sides of the triangle at the feet of the altitudes of the triangles, as if a ray of light bounced the sides of the triangle to the point of beginning. Necessarily, $CD_1=C'D_2$ . The blue line $D_1 D_2$ is shortest when $AD_1=AD_2$ is shortest, or the base of the altitude of triange $ABC$ . The dotted blue line (which does not have minimum perimeter) also makes equal angles with the triangle except where it touches side $BC$ . Only the orthic triangle makes equal angles with the triangle at all three vertices. Please see John Ross' elegant solution to this, based on the similarity of smaller triangles with triangle $ABC$ .

The dimensions of the triangle has been altered for clarity.

One can instead make this a calculus problem by finding the mininum length of the blue line, given the condition that it must start and end in the same place, from $D_1$ to $D_2$ , involving a single variable proportional to $CD_1$ . Let's try that.

Given triangle with sides $(9,5,10)$ as shown above in the problem statement, with the left bottom vertex at $(0,0)$ . $\;$ Let $k=\dfrac{CD_1}{CB'}=\dfrac{C'D_2}{C'B}$ . $\;$ Then the coordinates of the top and the bottom ends of the blue line are

$\left( \dfrac{3}{125}(325+347k), \dfrac{6\sqrt{14}}{125}(25+19k) \right)$
$\left( \dfrac{39}{5}(1-k), -\dfrac{6\sqrt{14}}{5}(1-k) \right)$

so that the mininum distance is $8.96$ at $k=\dfrac{1}{27}$ , and the blue line passes through the feet of altitudes as shown.

This solution is by no means elegant or even that mathematical, but if this problem was to be solved in under 30 seconds here is how to do it. First, let's consider some easy triangle like an isosceles triangle and let's imagine we had to find the smallest possible perimeter of the inscribed triangle. Intuitively, it would look something like this: Where $\varepsilon$ is some tiny distance, so we can say $l_1=l_2$ . Since we are minimizing distance $l_1$ and $l_2$ are perpendicular to the line on the base of the triangle. Now finding the smallest perimeter is merely calculating 2 $\times$ $l_1$ or 2 $\times$ $l_2$ (not that it matters). Now let's consider the triangle given: The line $Y$ seems to be the shortest (one could check, but we will proceed assuming $Y$ is the shortest). Since $Y\perp AB$ we can form a right triangle at $\angle ABC$ where the side $Y$ multiplied by $2$ will give the smallest perimeter. First to find $\angle ABC$ we use the cosine law.

$\angle ABC = \cos^{-1}(\frac{AC^2-AB^2-BC^2}{-2(AB)(BC)}) = \cos^{-1}(\frac{9^2-10^2-5^2}{-2(10)5)}) = 63.896^\circ$

From here, we use sin ratio (since $Y\perp AB$ ) to calculate the distance from $AB$ to the vertex $C$ which is $Y$

$Y = 5 \times \sin{63.896^\circ}=4.4899$

So the smallest perimeter is $P = 2(Y) = 2(4.4899) = 8.97$ (yes this is slightly off, but much easier)

It is very interesting to think about this problem in a way of physics. Suppose the three sides of the given triangle are three rigid rods with zero friction. Then we put a elastic band around the rods, and it will surely try to shrink to a minimum length. And by analyzing the forces in the final balance, the conclusion of orthic triangle will be obvious.

Calculated the (x,y) cordinates of the given triangle and it's orthic triangle and the perimeter= (13/3)+(2/3) +(99/25) = 8.96,

Answer=8.96

I solved the problem by using analytic geometry concepts, setting intersection of lines 9, and 10 as (0,0) and setting intersection of lines 5 and 10 as (10,0). Then, I solved for the point of intersection of lines 9 and 5 using the concepts of distance formula. I solved it to be (7.8, 4.4899).

I generalized a certain triangle with vertices as lying in the 3 lines. Thus, I expressed perimeter as an equation dependent on 3 variables.In this equation, it is composed of 3 addends: length of arbitrary point from 9 to 10, length of arbitrary point from 10 to 5, length of arbitrary point from 5 to 9.

Then, I created a Python code and used exhaustive enumeration with 0.01 increments in substituting it in the equation. I generated the minimum perimeter as 8.979977728257978.

I solved this for 2 hours, (1 and a half hour for the written solution, 30 minutes for the Python code). Wew. Got it correct, I guess.

Seemed like if it had been an quilateral triangle, the answer would have been “half the perimeter of the original” but 12 was wrong, so I figured it was a related number. I guessed 9 and the app rounded down. I don’t really think this was a clever solution, so why I’m really here just asking why it worked well enough?

To minimize the perimeter of this triangle, it is best to use points around the incenter of the triangle. This means that the triangle we seek is formed by the triangle's 3 points of tangency with it's incircle. Using a bit (a lot) of cheating from https://www.triangle-calculator.com/?what=sss&a=9&b=5&c=10&submit=Solve , I got this graph:

https://www.desmos.com/calculator/in5bn70a9y

Then, by finishing with the Pythagorean Theorem and adding up all the lengths of the triangle we get $\boxed{8.96}$ .

I guessed and got it right on my second try

Although the more precise answer appears to be 8.96, my answer of 8.981 was also accepted as correct.

To arrive at this answer, I first determined that the smallest internal triangle will be an infinitely acute triangle with its 2 longest sides running from the longest side of the outer triangle (the side of length 10) to two points infinitely close to the outer triangle's opposite vertex (the vertex connecting sides of length 9 and 5).

I calculated the height of the triangle (the distance between that vertex and the side of length 10) to be 4.49. I then accepted 4.49 as the length of the 2 longest sides of the inter triangle.

With just 3 significant digits to work with, I estimated that the length of the inner triangle's 3rd side (connecting the outer triangle's sides of length 9 and 5) to be .001.

Addition to find the perimeter of the inner triangle: 4.49 + 4.49 + .001 = 8.981.

My conclusion: Given the precision in the other solutions, I'm surprised that 8.981 is an acceptable answer. And at the same time, despite the precision of the other solutions, I also don't understand how the perimeter of the inner triangle can be smaller than 8.98 (or twice the height of the outer triangle, or 2 * 4.49).

Let theta = the angle included between the sides of length 9 and 5. By the Law of Cosines, cos theta = 1/15 and sin theta = (4/15)sqrt 14. Let C be the vertex of theta and let F be the foot of the altitude from C. By Heron's formula the area of the triangle is 6sqrt 14, CF = (6/5)sqrt 14. Let X and Y be images of F upon reflection in the sides of length 9 and 5 respectively. Since CX = CF = CY, triangle XCY is isosceles with base XY equal to the perimeter, P, of the inscribed triangle of minimum perimeter (orthic triangle))...Why? Now notice that angle XCY = 2theta, so dividing triangle XCY into two congruent right triangles yields the equation (XY/2) / CX = (XY/2) / CF = XY/(2CF) = sin theta, or equivalently, XY/ ((12/5)sqrt 14) = (4/15)sqrt 14. Therefore, the perimeter P of the inscribed triangle of minimal perimeter = XY = 224/25 = 8.96.