In a triangle $ABC$, the base $AB = 6 \mbox{cm}$. The vertex $C$ varies such that the area is always equal to $12 \mbox{cm}^2$. Find the minimum value of the sum $CA+CB$.
My attempt: set $AB$ as a $base = 6$ cm; after that I apply the area of triangle formula that is $area = (\frac{1}{2}) base* height = 12 cm^2$ and i got $height = 4$ cm. From that I find Right angle triangle ..I again apply the Pythagorean theorem .I got the value... $CB + CA = \sqrt{ 52} + 6$. Is my answer is correct or not?
On
I think your way is not true. Why the minimum occurs for right-angled triangle?
I think we can solve your problem by the following way. Let $CD$ be an altitude of the triangle and $BD=x$.
Thus, $$CD=\frac{2\cdot12}{6}=4$$ and by Minkowski (triangle inequality) we obtain: $$AC+BC=\sqrt{(6-x)^2+4^2}+\sqrt{x^2+4^2}\geq\sqrt{(6-x+x)^2+(4+4)^2}=10.$$ The equality occurs for $(6-x,4)||(x,4)$, which happens for $x=3$.
Id est, we got a minimal value.
Done!
On
Let's consider this triangle on the coordinate plane. Let A has coordinates (0,0) and B has coordinates (6,0). Point C will belong either to line y=4 or y=-4 but it's irrelevant, both cases will produce the same result. Let x be the abscissa of point C. We can define CA+CB as the following function $f(x)=\sqrt{x^2+4}+\sqrt{(6-x)^2+4}$. Now all we need to do is to find the minimum of this function when $x$ is between 0 and 6. $\frac{df}{dx}=\frac{x}{\sqrt{x^2+4}} - \frac{x-6}{\sqrt{(6-x)^2+4}}$. Solving $\frac{df}{dx}=0$ we find that $x$=3. Now all we need to do is to check the values of $f(0), f(3), f(6)$ to prove that the minimal value is $f(3)$.
The locus of $C$ is two lines parallel to $AB$, one line lie above $AB$ by 4 cm, one line lie below $AB$ by 4 cm. We denote the upper line by $l$. We only need to consider the case when $C$ lies on $l$.
Let $B'$ be the reflection of $B$ along $l$, then $CA+CB = CA+CB'$, so $CA+CB$ is minimized when $C,A,B'$ are collinear, which implies $CA=CB$, the triangle is isosceles in this case.