Suppose that $A B C D$ is a square with side length 4 and that $0<k<4$. Let points $P, Q, R$, and $S$ be on $B C, C D, D A$, and $A P$, respectively, so that

$$
\frac{B P}{P C}=\frac{C Q}{Q D}=\frac{D R}{R A}=\frac{A S}{S P}=\frac{k}{4-k}
$$

What is the value of $k$ which minimizes the area of quadrilateral $P Q R S$ ?

![](https://cdn.mathpix.com/cropped/2024_04_17_bceb232fd49f7f0368beg-2.jpg?height=330&width=333&top_left_y=2082&top_left_x=1357)

## PART B

For each question in Part B, your solution must be well-organized and contain words of explanation or justification. Marks are awarded for completeness, clarity, and style of presentation. A correct solution, poorly presented, will not earn full marks.
Since square $A B C D$ has side length 4 , then its area is $4^{2}$, which equals 16 .

The area of quadrilateral $P Q R S$, which we expect to be a function of $k$, equals the area of square $A B C D$ minus the combined areas of $\triangle A B P, \triangle P C Q, \triangle Q D R$, and $\triangle A R S$.

Since $\frac{B P}{P C}=\frac{k}{4-k}$, then there is a real number $x$ with $B P=k x$ and $P C=(4-k) x$.

Since $B P+P C=B C=4$, then $k x+(4-k) x=4$ and so $4 x=4$ or $x=1$.

Thus, $B P=k$ and $P C=4-k$.

Similarly, $C Q=D R=k$ and $Q D=R A=4-k$.

$\triangle A B P$ is right-angled at $B$ and so its area is $\frac{1}{2}(A B)(B P)=\frac{1}{2}(4 k)=2 k$.

$\triangle P C Q$ is right-angled at $C$ and so its area is $\frac{1}{2}(P C)(C Q)=\frac{1}{2}(4-k) k$.

$\triangle Q D R$ is right-angled at $D$ and so its area is $\frac{1}{2}(Q D)(D R)=\frac{1}{2}(4-k) k$.

To find the area of $\triangle A R S$, we first join $R$ to $P$.

![](https://cdn.mathpix.com/cropped/2024_04_30_2d11aa317d3213e95c24g-04.jpg?height=331&width=331&top_left_y=886&top_left_x=954)

Now $\triangle A R P$ can be seen as having base $R A=4-k$ and perpendicular height equal to the distance between the parallel lines $C B$ and $D A$, which equals 4 .

Thus, the area of $\triangle A R P$ is $\frac{1}{2}(4-k)(4)$.

Now we consider $\triangle A R P$ as having base $A P$ divided by point $S$ in the ratio $k:(4-k)$.

This means that the ratio of $A S: A P$ equals $k:((4-k)+k)$ which equals $k: 4$.

This means that the area of $\triangle A R S$ is equal to $\frac{k}{4}$ times the area of $\triangle A R P$. (The two triangles have the same height - the distance from $R$ to $A P$ - and so the ratio of their areas equals the ratio of their bases.)

Thus, the area of $\triangle A R S$ equals $\frac{\frac{1}{2}(4-k)(4) \cdot k}{4}=\frac{1}{2} k(4-k)$.

Thus, the area of quadrilateral $P Q R S$ is

$$
\begin{aligned}
16-2 k-3 \cdot \frac{1}{2} k(4-k) & =16-2 k-\frac{3}{2} \cdot 4 k+\frac{3}{2} k^{2} \\
& =\frac{3}{2} k^{2}-2 k-6 k+16 \\
& =\frac{3}{2} k^{2}-8 k+16
\end{aligned}
$$

The minimum value of the quadratic function $f(t)=a t^{2}+b t+c$ with $a>0$ occurs when $t=-\frac{b}{2 a}$ and so the minimum value of $\frac{3}{2} k^{2}-8 k+16$ occurs when $k=-\frac{-8}{2(3 / 2)}=\frac{8}{3}$.

Therefore, the area of quadrilateral $P Q R S$ is minimized when $k=\frac{8}{3}$.

ANSWER: $k=\frac{8}{3}$

## Part B