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This problem comes from a Grade 11 math textbook in a section on solving quadratic functions by factoring. What follows is my attempt at a solution.

A right triangle has one leg, $n$ and a hypotenuse, $2n+6$. The challenge is to determine the lengths of all the sides, given that the perimeter is $60$.

I worked through this problem a few times with a student. Finally I suggested we step away from it. I decided to post it to this forum for everyone's feedback. As I progressed through the problem, a strange thing happened -- a solution emerged. As is often the case, slowing down and meticulously formulating a question is all that is needed to find a solution.

Rather than include my attempt in the question body, I've added it as a solution below.

Blue
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MathAdam
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3 Answers3

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Let's correct your solution. Following your notation, the other leg will have length $x$. Since the perimeter is $60$, \begin{align*} x + n + 2n + 6 & = 60\\ x + 3n + 6 & = 60\\ x & = 54 - 3n \end{align*} By the Pythagorean Theorem, \begin{align*} x^2 + n^2 & = (2n + 6)^2\\ x^2 + n^2 & = 4(n + 3)^2\\ x^2 + n^2 & = 4(n^2 + 6n + 9)\\ x^2 + n^2 & = 4n^2 + 24n + 36\\ x^2 & = 3n^2 + 24n + 36\\ x^2 & =3(n^2 + 8n + 12) \end{align*} On the other hand, we know that \begin{align*} x^2 & = (54 - 3n)^2\\ x^2 & = 9(18 - n)^2\\ x^2 & = 9(324 - 36n + n^2)\\ \end{align*} Hence, \begin{align*} 3(n^2 + 8n + 12) & = 9(n^2 - 36n + 324)\\ n^2 + 8n + 12 & = 3(n^2 - 36n + 324)\\ n^2 + 8n + 12 & = 3n^2 - 108n + 972\\ 0 & = 2n^2 - 116n + 960\\ 0 & = n^2 - 58n + 480\\ 0 & = (n - 10)(n - 48) \end{align*} giving $n = 10$ and $n = 48$ as possible solutions.

If $n = 10$, then $2n + 6 = 26$, and $x = 54 - 3n = 24$. As you can check, $(10, 24, 26)$ is a Pythagorean triple.

If $n = 48$, then $2n + 6 = 102$, and $x = 54 - 3n = -90$. Since a side length cannot be negative, we reject this solution.

You confused the roles of $x$ and $n$, which led you astray.

N. F. Taussig
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  • Thank you. I had begun to edit the answer to correct the $x$ vs. $n$ confusion. I changed it back after I saw your solution and correction. Keeping the error in my answer helps preserve the flow (and mis-flow) of thought and increases the didactive value of this question. Thanks for setting me straight! – MathAdam Jun 02 '17 at 13:29
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I've designated the other leg of the triangle, $x$.

  1. Given the perimeter, $$60=2n+6+n+x$$ $$54=3n+x$$ $$x=54-3n=-3(n-18)$$ $$x^2=9(n^2-36n+324)$$
  2. By Pythagoras, $$x^2+n^2=(2n+6)^2=4n^2+24n+36$$ $$x^2=3n^2+24n+36$$ $$=3(n^2+8n+12)$$
  3. Combining the above two results, $$=3(n^2+8n+12)=9(n^2-36n+324)$$ $$=(n^2+8n+12)=3(n^2-36n+324)$$ $$=3n^2-108n+972$$ $$0=2n^2-116n+960=n^2-58n+480$$ $$=(x-48)(x-10)$$

This leaves two so!utions, $x={10,48}.$ From here, $$54=3n+x_1 = 3n+10$$ $$n=\frac{44}{3}$$ $$2n+6 = \frac{106}{3}$$ Alternatively,  $$54=3n+x_2 = 3n+48$$ $$n=2$$ $$2n+6 = 10$$ The solutions for the sides are $$  10, \frac{44}{3},  \frac{106}{3}$$ and $$48,2,10$$

MathAdam
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Let the blue point be $n$ units above the origin, $n > 0$.

Construct a circle centred at the blue point with radius $2n+6$. Then by the intersecting chords theorem, by choosing the $x$- and $y$-axis as chords, $(2n+6+n)(2n+6-n) = x^2$. Hence the length of the other leg is $\sqrt{(3n+6)(n+6)} = \sqrt{3n^2 + 24n + 36}$.

Thus $n + 2n + 6 + \sqrt{3n^2 + 24n + 36} = 60$, then squaring both sides:

$$3n^2 + 24n + 36 = (60 - 3n - 6)^2$$ $$3n^2 + 24n + 36 = 2916 - 324n + 9n^2$$ $$6n^2 - 348n + 2880 = 0$$ $$n^2 - 58n + 480 = 0$$ $$(n - 10)(n - 48) = 0 \implies n = 10, 48$$

but when $n = 48$, the hypotenuse $2(48) + 6 > 60$, which is the perimeter. Hence only $n= 10$ is possible, which yields a $10-24-26$ right triangle.

Toby Mak
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