# Thread: min. area of triangle

1. ## min. area of triangle

p228 q20c
given area of triangle $\displaystyle A = \frac{2(r^2+r+1)}{(1+r)^2}$
find the min. area
i put r --> infinty then A = 2 but it is wrong. the answer of my book says A=3/2

2. Originally Posted by afeasfaerw23231233
p228 q20c
given area of triangle $\displaystyle A = \frac{2(r^2+r+1)}{(1+r)^2}$
find the min. area
i put r --> infinty then A = 2 but it is wrong. the answer of my book says A=3/2

$\displaystyle A = 2\left(1 - \frac{r}{(1+r)^2}\right)$

A achieves minimum when $\displaystyle \frac{r}{(1+r)^2}$ is maximum.

But $\displaystyle (1+r)^2 \geq 4r \Rightarrow \frac{r}{(1+r)^2} \leq \frac14 \Rightarrow \left(1 - \frac{r}{(1+r)^2}\right) \geq \frac34 \Rightarrow A \geq \frac32$

With equality at r=1.

So minimum of A is achieved at r=1.And that is A = 3/2.

3. how to prove
$\displaystyle (1+r)^2 \geq 4r$
if it is $\displaystyle (k+r)^2$ , then $\displaystyle (k+r)^2 \ge ?$

4. Originally Posted by afeasfaerw23231233
how to prove
$\displaystyle (1+r)^2 \geq 4r$
[snip]
To prove that $\displaystyle 1 + 2r + r^2 \geq 4r$ you only have to prove that $\displaystyle 1 + r^2 \geq 2r$.

Draw a graph of $\displaystyle y = x^2 + 1$ and $\displaystyle y = 2x$ and it's not hard to see (and then prove) that the line $\displaystyle y = 2x$ is tangent to $\displaystyle y = x^2 + 1$ at (1, 2) ........

5. Originally Posted by Isomorphism
[tex][snip]when $\displaystyle \frac{r}{(1+r)^2}$ is maximum.

[snip]at r=1.
[snip]
This result can also be easily proved using calculus.

6. @mr fantastic: $\displaystyle 1+r^2\ge 2r \iff r^2-2r+1\ge 0 \iff \left(r-1\right)^2\ge 0 \,\Box$.