# Find the perimeter of the triangle

#### anemone

##### MHB POTW Director
Staff member
The two shortest sides of a right-angled triangle, $a$ and $b$ satisfy the inequality $$\displaystyle \sqrt{a^2-6a\sqrt{2}+19}+\sqrt{b^2-4b\sqrt{3}+16}\le3$$.

Find the perimeter of this triangle.

##### Well-known member
The two shortest sides of a right-angled triangle, $a$ and $b$ satisfy the inequality $$\displaystyle \sqrt{a^2-6a\sqrt{2}+19}+\sqrt{b^2-4b\sqrt{3}+16}\le3$$.

Find the perimeter of this triangle.
=> sqrt((a - 3 sqrt(2))^2 + 1) + sqrt((b - 2 sqrt(3))^2 + 4) < = 3

possible onlly if a = 3 sqrt(2) and b = 2 sqrt(3)

so c(diagonal) = sqrt(30)

so perimeter = 3 sqrt(2) + 2 sqrt(3) + sqrt(30)

- - - Updated - - -

#### anemone

##### MHB POTW Director
Staff member
=> sqrt((a - 3 sqrt(2))^2 + 1) + sqrt((b - 2 sqrt(3))^2 + 4) < = 3

possible onlly if a = 3 sqrt(2) and b = 2 sqrt(3)

Thanks for participating and I think it's necessary to state why the given inequality is true iff $a = 3 \sqrt{2}$ and $b = 2 \sqrt{3}$, what do you think?

Or perhaps it's very obvious and it's just me don't see how it is so? ##### Well-known member

Thanks for participating and I think it's necessary to state why the given inequality is true iff $a = 3 \sqrt{2}$ and $b = 2 \sqrt{3}$, what do you think?

Or perhaps it's very obvious and it's just me don't see how it is so? I think I owe an explanation
sqrt((a - 3 sqrt(2))^2 + 1) + sqrt((b - 2 sqrt(3))^2 + 4) < = 3

now we are having

sqrt(1 + x) + sqrt(4 + y) <= 3 with x,y > 0

if x = 0 and y =0 then LHS = 3
if x > 0 then y = 0 then LHS = 2 + sqrt(1+x) > 2 + 1 > 3
similarly for y > 0 and for x and y > 0 LHS > 3

x = (a - 3 sqrt(2))^2
y = (b - 2 sqrt(3))^2

or

lowest value of LHS = 3 when - (a - 3 sqrt(2))= 0 and (b - 2 sqrt(3)) = 0 then only condition is satisfied