Is there a triangle with $12 \, cm^2$ area and $12$ cm perimeter?
Problem
Source: jbmo shortlist 2003
Tags: geometry, JBMO, area
mathdragon2000
12.10.2017 22:04
Answer: No.
First part of sol: Let the side lengths be $ a, b, c $. Then $ a + b + c = 12 $ and $ \sqrt{\frac{a + b + c}{2} \cdot \frac{b + c - a}{2} \cdot \frac{a - b + c}{2} \cdot \frac{a + b - c}{2}} = 12 $ by Heron's Formula. We have $$ \frac{(a + b + c) - 2a}{2} \cdot \frac{(a + b + c) - 2b}{2} \cdot \frac{(a + b + c) - 2c}{2} = 24 \implies $$$$ (6 - a)(6 - b)(6 - c) = 24 $$
RagvaloD
12.10.2017 22:18
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
scrabbler94
12.10.2017 23:21
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.
satya25
20.06.2019 17:27
........
potatothegeek
20.06.2019 17:31
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 $6 cm^2$.
minageus
20.06.2019 18:53
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.
Mixer_V
20.06.2019 18:55
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.
TYT
20.06.2019 19:08
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: \[ A \le \frac{P^2}{12\sqrt{3}} \]
Delta0001
20.06.2019 19:10
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
Claim: Equilateral triangle has maximum area with given perimeter
Heron's formula gives the area of a triangle with sides $a$, $b$, and $c$ as $A = \sqrt{s(s-a)(s-b)(s-c)}$, where $s=\displaystyle\frac{1}{2}(a+b+c)$.
Let $2s=a+b+c>0$, and suppose $A^2 = s(s-a)(s-b)(s-c)$ is maximized for $a$, $b$, and $c$ that are not all equal. Without loss of generality, assume $a<b$.
It's straightforward to show that $s(s-\displaystyle\frac{a+b}{2})(s-\frac{a+b}{2})(s-c)>A^2$, which means that you'll get a triangle of larger area if you replace the unequal sides $a$ and $b$ with equal sides, keeping the same perimeter. This contradicts the assumption that $a\neq b$ at the maximum.
Main Problem:
Area of an equilateral triangle $A = \frac{\{sqrt{3}}{4} a^2$ and perimeter $P = 3a$. Hence, $A \le \frac{P^2 \sqrt{3}}{36} ~ \blacksquare$
Darn sniped
minageus
20.06.2019 21:05
Ok thank you very much!
ATGY
23.07.2020 13:56
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 $12$
OlympusHero
25.07.2021 19:36
lol
Given a perimeter, the area of a triangle is maximized when it is equilateral (well-known) so the maximum possible area is that of an equilateral triangle with side length $4$. This makes an area of $\frac{4^2\sqrt3}{4}=4\sqrt3$, but this is smaller than $12$. This means our answer is no.
Math_legendno12
24.06.2023 12:45
(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
mathmax12
23.07.2023 17:56
The area is maximized when equalateral triangle. So the side length of $4$ gives area of $4\sqrt{3} < 12$, so it is impossible.