GRE study guide asks
The perimeter of square S is 40. Square T is inscribed in square S. What is the least possible area of square T?
Choices are
They say answer is 50. How do they even get this? T has lengths less than 10, so it can be 1 x 1 square, or 9 x 9 square. Please guide.
On
The definition of an "inscribed" square in a square is that all of the smaller square's vertices lies on the boundaries of the larger square.
Notice that for this to happen, the smaller square's vertices will divide each side of the larger square into two segments, the same on each side. Let's call their lengths $a$ and $b.$ We have that $a + b = 10.$ The side length of the smaller square is, by Pythagoras, $s = \sqrt{a^{2} + b^{2}}.$ Then the area of the smaller square must be $s^{2} = a^{2} + b^{2}.$
You can use calculus here, but common sense also tells us that the minimum occurs when $a = b = \frac{10}{2} = 5.$ So the minimum area of the smaller square is $5^{2} + 5^{2} = \boxed{50}.$ This is answer choice $4.$
If it is inscribed it means that all vertices of $T$ lie on $S$. Like this
This divides the square into 4 triangle pieces and one square piece. We need to find the hypotenuse of the triangle piece to find the area of the inner square. The triangle has 2 sides of length 5. Using the pythagorean theorem yields $h = \sqrt(5^2 + 5^2) = \sqrt{50}$. The area of a square with side length $h$ is $h^2$ so the area of the inner square is $\sqrt{50}^2 = 50$. Now you may wonder why the inscription has to be like this to yield a minimum. Lets consider the general case.
The area of the inner square is minimized when the area of the outer triangles is maximized. You can consider the vertices of the inner square dividing the sides of the outer square into two pieces. In the optimal case the length of these pieces are the same. This intuitively seems in the first diagram we can prove this to be true in general.
Let $L$ be the side length of the square. Say the vertices divides the side into pieces of length $a$ and $b$. We know that $a + b = L$ so $b = L - a$. Notice that all the triangles have the same area. So we only need to minimize one. The area of the triangle, $A$, will be $A = \frac{1}{2}ab = \frac{1}{2}a(L - a)$. We can find the minimum without calculus because this is a parabola. We just need to find the vertex of the parabola. Instead of completing the square, we can use the roots of the parabola as we have it in factored form. Recall that if the parabola has roots $r_1$ and $r_2$ then its vertex has an $x$ coordinate of $\frac{r_1 + r_2}{2}$(Note that we dont care about the actual value of the maximum area for this proof, but rather the value of $a$ that maximizes the area of the triangle). Since our quadratic has roots $0$ and $L$, then the x coordinate of the vertex is $\frac{0 + L}{2} = \frac{L}{2}$. Therefore the values of $a$ and $b$ that maximize the triangle area is $a = \frac{L}{2}$, $b = L - \frac{L}{2} = \frac{L}{2}$. This yields the result that the minimum case has $a = b$.