Sides of a triangle form an arithmetic sequence with common difference $2$, and its area is $6 \text{ cm }^2$. Find its sides.
Problem
Source: Albania Round 2, Grade XI 2014 P2
Tags: algebra, Sequences, Arithmetic Progression
PhysicsMonster_01
15.11.2018 10:05
@below it'll be better(easier calculations) if you assume the sides to be $x-2, x, x+2$.
Bowser498
15.11.2018 10:19
Let the sides be $x, x+2, x+4$. The best thing to do is to use Heron's formula. The semiperimeter is $\frac{3x+6}{2}$, so $6 = \sqrt{(\frac{3x+6}{2})(\frac{3x+6}{2}-x)(\frac{3x+6}{2}-(x+2))(\frac{3x+6}{2}-(x+4))} \rightarrow 36=\frac{3}{16}x^{4}+\frac{3}{2}x^{3}+\frac{3}{2}x^{2}-6x+9 \rightarrow \frac{3}{16}x^{4}+\frac{3}{2}x^{3}+\frac{3}{2}x^{2}-6x-27=0$. We get that the only positive solution to the equation is $2\sqrt{6}-2$ so the three sides are $2\sqrt{6}-2$, $2\sqrt{6}$, and $2\sqrt{6}+2$
gautamlucky2002
15.11.2018 11:37
Let the sides of the triangle be $x-2,x,x+2$. Then, the semi perimeter ($s$) would be equal to $\frac{3x}{2}$ . From Heron's Formula, area of the triangle will be equal to $\sqrt{(\frac{3x}{2})(\frac{x}{2})(\frac{x}{2}-2)(\frac{x}{2}+2)}$ which is equal to $6 cm^2$. On solving, we would get
$\frac{x}{4}\sqrt{3x^2-48} = 6$. On further solving, we would get $x^2(x^2-16)=192$. Let $x^2=y$ then, $y^2-16y-192=0$. On solving, we would get
$y=24$ or $y=-8$ . Since $y=x^2$, y will always be positive. Hence, $y=24$. Hence, $x^2=24$, giving $x=2\sqrt{6}$, as $x$ is always positive ($x$ is a side of a triangle). With this, the sides of the triangle are $2\sqrt{6}-2,2\sqrt{6},2\sqrt{6}+2$.
TheCosineLaw
15.11.2018 13:37
Shouldn't this be in Middle School Math? This is way to easy.
AopsUser101
28.10.2019 18:17
TheCosineLaw wrote: Shouldn't this be in Middle School Math? This is way to easy. This came from an Albania Math Olympiad Contest, so even though this problem is mid/late AMC, early AIME #1-#2 level, it still belongs in this forum. SOLUTION Let the sides of the triangle be $a-2,a,a+2$ and the semiperimeter $\frac{3a}{2}$. Then, by Heron's formula, we have that the area would be$\sqrt{\left( \frac{3a}{2}\right)\left( \frac{a+4}{2}\right)\left( \frac{a}{2}\right)\left( \frac{a-4}{2}\right)}=6$. Squaring and cross-multiplying, we see that $3a(a+4)(a)(a-4)=36 \cdot 16 \implies$ $3a^2(a^2-16)=36 \cdot 16 \implies$ $a^4-16a^2-192=0$. By either factoring or using the quadratic formula, we see that $a^2=24$, so $a=2\sqrt{6}$. The side lengths of the triangle are therefore $2\sqrt{6}-2,2\sqrt{6},$ and $2\sqrt{6}+2$.
OlympusHero
20.10.2021 00:15
Let the sides be $2a-2, 2a, 2a+2$. Then the area is $\sqrt{(3a)(a+2)(a)(a-2)}=6$, so $(3a^2)(a^2-4)=36$. Denote $a^2=x$, so $3x(x-4)=36 \implies x=6$. Then $a=\sqrt6$, so the answer is $2\sqrt6-2, 2\sqrt6, 2\sqrt6+2$.
Mr.Sharkman
15.07.2022 21:54
WE can easily do this with herons formula.
mathmax12
05.09.2023 03:03
Let, the sides, be $a-2, a, a+2,$ then, by Heron's $\frac{3a}{2}\cdot \frac{a}{2}\cdot \frac{a+4}{2}\cdot \frac{a-4}{2}=36$, hence, $3a(a+4)a(a-4)=36\cdot 16$, hence, $a^4-16a^2-192$, now, let $a^2=b$, solving for $b,$ we get that, $b=24$, hence, $a=2\sqrt{6}$, so the side lengnths, are $2\sqrt{6}-2, 2\sqrt{6}, 2\sqrt{6}+2.$