# Does there exist a right triangle with area 7 and perimeter 12?

This question is really trivial. I can prove that there is no right triangle with area 7 and perimeter 12, but what I do is solve the following system: if $a$, $b$ and $c$ are, respectively, the two legs and hypotenuse of such a triangle, then

$$a^2 + b^2 = c^2,$$ $$a + b +c = 12,$$ $$ab = 14$$

It is easy (although a bit boring and long) to see that there are no real solutions to this system.

But I feel that there is a simple answer to this question - perhaps using the triangle inequality - but I just cannot see it.

• "it is easy". Citation needed Aug 17, 2018 at 15:36

I'm not sure if this counts as a "simple answer", but let $x$ be the length of one leg of a right triangle of area $7$; then the other leg is $\frac{14}x$, and the hypotenuse is $\sqrt{x^2 + \left(\frac{14}x\right)^2}$, so the perimeter is given by the function $$P(x) = x + \frac{14}x + \sqrt{x^2 + \left(\frac{14}x\right)^2}$$

Asymptotically, we have $\lim_{x\to 0} P(x) = +\infty$ and $\lim_{x\to\infty} P(x) = +\infty$, and intuitively it seems clear that the absolute minimum occurs when the two legs of the triangle are equal in length, i.e. when $x=\sqrt{14}$. At this value, we have $$P(\sqrt{14}) = 2\sqrt{14} +2\sqrt{7}$$ This is the smallest possible perimeter for a right triangle of area $7$. It remains only to convince yourself that this number is greater than $12$.

If the perimeter is $12$, the triangle of maximum area is the equilateral triangle of side $4$ which has area $2\times 2\sqrt 3\lt 7$ (square both sides $48\lt 49$)

Now it is a question of how you show that the triangle of maximum area is equilateral. First fix a base AB. The third point C lies on an ellipse, and symmetry/convexity show that the maximum area is if the two remaining sides AC and BC are equal. So if the sides are not equal, you will get a greater area by averaging the shortest and longest. For maximum area (no further gains) all sides must be equal - hence equilateral.

In summary, a more formal way of showing that the maximal area of a triangle of fixed perimeter is reached when the triangle is equilateral is to go via Heron's formula (the area of a triangle as a function of its sides) and AM/GM. With the perimeter $2s=a+b+c$ fixed, note that $(-a+b+c)+(a-b+c)+(a+b-c)=2s$ and the product of the three terms in brackets is therefore greatest when they are equal.

Heron's formula is area =$\sqrt {s(s-a)(s-b)(s-c)}$

• This is a much better answer than mine. Finding the maximum area for a given perimeter is much more straightforward than finding the minimum perimeter for a given area. :) Aug 17, 2018 at 17:56
• @mweiss It's horses for courses - here the bound is good enough, which is fine. But there are cases for which this approach would not work and a more detailed analysis would be necessary. You can't always tell easily in advance. Aug 17, 2018 at 18:29
• Showing that shortening the base and lengthening the other two sides (or vice versa) does not increase area is a bit of a pain. You have to show that any such transformation of side length (keeping the sum intact) can be reversed by successivly fixing a base and transferring length from the shorter side to a longer side of the remaining 2 sides (which I believe is true, just annoying to prove). Or is there an easier way? (your argument shows that 2-side manipulation cannot increase area unless they move towards equal; showing it for 3 is my issue)
– Yakk
Aug 17, 2018 at 18:33
• @Yakk You don't need to do this. The dynamics here are if the three sides are not equal you can choose the middle side as base an increase the area by equalising the others. So no non-equilateral triangle can have maximum area for given perimeter. This method of optimising a three parameter problem by fixing one and showing that the others need to be equal, and concluding that all are equal at the critical point, is not particularly elegant in general, but works more often than it ought. Heron and AM/GM will do it "properly" here as noted. Aug 17, 2018 at 18:38

Note: This answer is to an earlier version of the question which asked for perimeter $14$.

What about \begin{align} a&=4+\sqrt2\\ b&=4-\sqrt2\\ c&=6 \end{align}

• Note you're answering the problem in the title, not the problem in the question.
– user14972
Aug 17, 2018 at 3:00
• @Hurkyl Not really. Look at the first sentence in the body of the question. The equation for the perimeter seems to be wrong. I pointed that out in a comment. Aug 17, 2018 at 3:02
• Oh gosh, my bad. There is indeed an error. Thank you Aug 17, 2018 at 3:03
• I got really confused seeing this answer. Then I realized the question was edited to replace perimeter 14 with 12. Aug 17, 2018 at 7:22
• @DavidRicherby Thanks for the edit. Aug 17, 2018 at 15:50

If the area was $$7$$ and the perimeter was $$12$$, the radius of the inscribed circle would be $$14/12$$ (exercise: why?). The circle is in the picture below.

On the other hand you can create any possible right triangle with perimeter $$12$$ by picking a point $$H$$ on the arc between $$U$$ and $$V$$ ($$T$$ is the center) and plotting a hypotenuse tangent to the arc at $$H$$ (exercise: why is the perimeter independent of $$H$$, if $$H$$ is on the arc?).

Imagine you move $$H$$ on the arc, generating all possible triangles. None of them would have the given circle as its inscribed circle. The radius of the circle is just to large (or the radius of the arc is too small). Note: This answer is to an earlier version of the question which asked for perimeter $14$.

There is indeed such a triangle.

The isosceles right triangle with legs $14/(2+\sqrt2)$ has perimeter $14$ and area

$$\frac12\left(\frac{14}{2+\sqrt2}\right)^2=147-98\sqrt2\approx8.4\;.$$

On the other hand, a right triangle with perimeter $14$ can be made to have arbitrarily small area by making one of the legs arbitrarily small. The right triangles with perimeter $14$ can be parametrized by the length of the shorter leg. The area is a continuous function of this parameter. By the intermediate value theorem, there is a right triangle with perimeter $14$ with some intermediate length of the shorter leg that has area $7$.

• Note you're answering the problem in the title, not the problem in the question.
– user14972
Aug 17, 2018 at 3:00
• @Hurkyl: See saulspatz's comment to your identical comment under this answer. Aug 17, 2018 at 3:30
• And see Lucas's response, comment on the OP, and updated question.
– user14972
Aug 17, 2018 at 3:31
• @Hurkyl: I do see that. I was merely pointing out that I answered the question that was at the time both the question in the title and the question in the text. Now that both have been edited, others can answer the edited question. Aug 17, 2018 at 3:33
• joriki, good answer - this was the tack I thought of. The only change to your answer I would suggest is to mention that the isosceles right triangle has the maximum area of any right triangle. you mentioned that shorter legs would have less area, but this would explain why you chose the isosceles to begin with.
– Jim
Aug 17, 2018 at 13:08

$$a+b=12-c$$ $$(a^2+b^2)+2ab=(a+b)^2=c^2-24c+144$$ $$c^2+28=c^2-24c+144$$ $$c=\frac{29}6$$ $$a+b=\frac{43}6\implies (a,b,c)=(k,\frac{43}6-k,\frac{29}6)$$ Since $c>a, c>b$, $$k\in(0, \frac{29}6)$$

• however, the found $a=k, b=\frac{43}{6}-k$ must satisfy $ab=14$, for which there will be no real $k$? it implies $c$ must be different... Aug 17, 2018 at 4:23
• That is the way I proved there is no right triangle satisfying the conditions, since $12 = ab = (46/6-k)k$ holds if and only if k is a complex number. Aug 17, 2018 at 8:08

Given a general triangle with inradius $r$, circumradius $R$ and semiperimeter $\rho=\tfrac12(a+b+c)$, the lengths of its sides $a,b,c$ can be found as three roots of the cubic equation

\begin{align} x^3-2\rho\,x^2+(\rho^2+r^2+4\,r\,R)\,x-4\,r\rho\,R &=0 \tag{1}\label{1} . \end{align}

For the right triangle the equation \eqref{1} can be simplified using a well-known condition

\begin{align} R&=\tfrac12(\rho-r) \tag{2}\label{2} \end{align} as follows:

\begin{align} (x+r-\rho)(x^2-(r+\rho)x+2r\rho) &=0 \tag{3}\label{3} . \end{align}

The first root, provided by the linear term in \eqref{3} is the size of hypotenuse,

\begin{align} c&=\rho-r , \end{align}

the other two sides must be

\begin{align} a,b&=\tfrac12\left(r+\rho\pm\sqrt{r^2-6r\rho+\rho^2}\right) . \end{align}

Or, in terms of the area $S$ and perimeter $p$,

\begin{align} c&=\tfrac12p-\tfrac{2S}{p} ,\\ a,b&= \frac{4S+p^2\pm\sqrt{p^4-24Sp^2+16S^2}}{4p} \tag{4}\label{4} . \end{align}

So, given $S=7$, $p=12$ we get

\begin{align} c&=\tfrac{29}6 ,\\ a,b&=\tfrac1{12}(43\pm\sqrt{167}\,i) , \end{align}

thus, the right triangle with declared properties is impossible.

Using \eqref{4} it is trivial to find out that the right triangle with $S=7$, $p=14$ has $c=6$, $a,b=4\pm\sqrt2$.

And as a bonus, the right triangle with $p=13$ is also valid and has side lengths

\begin{align} c&=\tfrac{141}{26}\approx 5.423 ,\\ a&=\tfrac1{52}(197-\sqrt{953})\approx 3.195 ,\\ b&=\tfrac1{52}(197+\sqrt{953})\approx 4.382 \end{align}

and is just slightly bigger than the famous $3-4-5$ right triangle.

Let me fill in a missing detail from the answer of @mweiss, namely, that a non-isosceles right triangle of perimeter $17$ does not have minimal area among all right triangles of perimeter $17$. And the specific value of perimeter does not matter for this purpose, so I'll do it for perimeter $p$.

Letting the legs of the triangle be $a,b$, from the Pythagorean theorem we get $$a^2 + b^2 = (p-a-b)^2$$ and simplifying we get $$2p a + 2p b - 2ab = (2p)^2$$ Differentiating implicitly with respect to $a$ we get $$2p + 2p \frac{db}{da} - 2b - 2 a \frac{db}{da} = 0$$ and so $$\frac{db}{da} = -\frac{p-b}{p-a} \qquad(*)$$ Since the area is $A=\frac{ab}{2}$ we also get $$2pa + 2pb - 4A = (2p)^2$$ Differentiating with respect to $a$ we get $$2p + 2p \frac{db}{da} - 4 \frac{dA}{da} = 0$$ Assuming $A$ is a minimum we get $$2p + 2p \frac{db}{da} = 0$$ and so $\frac{db}{da}=-1$. Combining this with $(*)$ and simplifying we get $$a=b$$

The triangle is two adjacent sides and one diagonal of a rectangle. As you pointed out, the product of the two legs is 14. This is the rectangle's area.

Making the rectangle a square minimises the sum of the triangle's legs and also minimises the rectangle's diagonal (the triangle's hypotenuse). This thus minimises the triangle's perimeter. Each leg is $\sqrt{14}$ and the hypotenuse is $\sqrt{28}$. This gives a perimeter $>12.7$. Thus no such triangle with perimeter 12 exists.