triangle with area = perimeter

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#1 h Oct 12, 2017, 6:57 PM • 2 Y

Y by Adventure10, Mango247

Is there a triangle with $12 \, cm^2$ area and $12$ cm perimeter?

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#2 h Oct 12, 2017, 7:04 PM • 2 Y

Y by Adventure10, Mango247

Partial solution

This post has been edited 1 time. Last edited by mathdragon2000, Oct 12, 2017, 7:07 PM

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#3 h Oct 12, 2017, 7:18 PM • 1 Y

Y by Adventure10

We can show, that for every triangle with area $A$ and perimeter $P$ is true that $A \leq \frac{P^2\sqrt{3}}{36}$ so answer is No

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#4 h Oct 12, 2017, 8:21 PM • 3 Y

Y by mathdragon2000, Adventure10, Mango247

Given a triangle with fixed perimeter, its area is maximized when the triangle is equilateral (same as what @above said). Then a triangle with perimeter 12 cm has maximum area $\frac{4^2 \sqrt{3}}{4} = 4 \sqrt{3}$ sq cm which is less than 12 sq cm, so answer is no.

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#5 h Jun 20, 2019, 2:27 PM • 2 Y

Y by Adventure10, Mango247

........

This post has been edited 4 times. Last edited by satya25, Jun 20, 2019, 2:34 PM
Reason: .

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#6 h Jun 20, 2019, 2:31 PM • 3 Y

Y by satya25, Adventure10, Mango247

satya25 wrote:

There is 1 triangle with such characters, it is be right angle triangle with hypotenuse 5cm and legs 3cm and 4cm.

No, a 3-4-5 triangle only has area $6 cm^2$ .

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#7 h Jun 20, 2019, 3:53 PM • 1 Y

Y by Adventure10

RagvaloD wrote:

We can show, that for every triangle with area $A$ and perimeter $P$ is true that $A \leq \frac{P^2\sqrt{3}}{36}$ so answer is No

How can we prove it? It seems similar to Weitzenbock's Inequality but I face difficulties in proving it. If someone can show me the way to solve this equality, I would be grateful.

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#8 h Jun 20, 2019, 3:55 PM • 1 Y

Y by Adventure10

minageus wrote:

RagvaloD wrote:

We can show, that for every triangle with area $A$ and perimeter $P$ is true that $A \leq \frac{P^2\sqrt{3}}{36}$ so answer is No

How can we prove it? It seems similar to Weitzenbock's Inequality but I face difficulties in proving it. If someone can show me the way to solve this equality, I would be grateful.

Show that the triangle with maximum area after you fixed a perimeter is the equilateral one, than just calculate his area and perimeter.

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TYT

#9 h Jun 20, 2019, 4:08 PM • 2 Y

Y by Adventure10, Mango247

Quote:

How can we prove it?

By Heron's formula we have:
$\begin{align*} A^2 &= \frac{1}{16} (a+b+c)(b+c-a)(c+a-b)(a+b-c) \\ &\le \frac{1}{16} (a+b+c) \times \frac{1}{27} (b+c-a + c+a - b + a + b -c)^3\\ &= \frac{1}{16} \times \frac{1}{27} (a+b+c)^4 = \frac{P^4}{16 \times 27}\\ \end{align*}$
So we finally arrive at:
$A \le \frac{P^2}{12\sqrt{3}}$

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#10 h Jun 20, 2019, 4:10 PM • 3 Y

Y by Pluto1708, Adventure10, Mango247

RagvaloD wrote:

We can show, that for every triangle with area $A$ and perimeter $P$ is true that $A \leq \frac{P^2\sqrt{3}}{36}$ so answer is No

Proof

Darn sniped :blush:

:blush:

This post has been edited 2 times. Last edited by Delta0001, Jun 20, 2019, 4:11 PM

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#11 h Jun 20, 2019, 6:05 PM • 2 Y

Y by Adventure10, Mango247

Ok thank you very much!

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#12 h Jul 23, 2020, 10:56 AM • 1 Y

Y by Mango247

The area of an triangle is maximized when given its perimeter when it is an equilateral (proof using heron's) triangle. Clearly, if the perimeter is $12$ , the largest possible area would be $\frac{16\sqrt3}{4} = 4\sqrt3$ which is clearly less than $12$

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#13 h Jul 25, 2021, 4:36 PM

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lol

Solution

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Math_legendno12

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Math_legendno12

#14 h Jun 24, 2023, 9:45 AM

Y by

(Without using equilateral triangle or heron's formula)

Choose an arbitrary side, length b and height h.
bh/2=Area=12
bh=24

12= perimeter
>b+2h
>= 2sqrt(2bh)
=2sqrt(48)
>12
Contradiction

So no triangles exist

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#15 h Jul 23, 2023, 2:56 PM

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The area is maximized when equalateral triangle. So the side length of $4$ gives area of $4\sqrt{3} < 12$ , so it is impossible.

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