Subfield: Geometry


Question: In the diagram, $D$ is the vertex of a parabola. The parabola cuts the $x$-axis at $A$ and at $C(4,0)$. The parabola cuts the $y$-axis at $B(0,-4)$. The area of $\triangle A B C$ is 4. Determine the area of $\triangle D B C$.

![](https://cdn.mathpix.com/cropped/2023_12_21_b301ea0eb47ddd66a1f3g-1.jpg?height=431&width=567&top_left_y=424&top_left_x=1256)

Solution:
First, we determine the coordinates of $A$.

The area of $\triangle A B C$ is 4 . We can think of $A C$ as its base, and its height being the distance from $B$ to the $x$-axis.

If the coordinates of $A$ are $(a, 0)$, then the base has length $4-a$ and the height is 4 .

Thus, $\frac{1}{2}(4-a)(4)=4$, so $4-a=2$ and so $a=2$.

Therefore, the coordinates of $A$ are $(2,0)$.

Next, we determine the equation of the parabola.

The parabola has $x$-intercepts 2 and 4 , so has equation $y=k(x-2)(x-4)$.

Since the parabola passes through $(0,-4)$ as well, then $-4=k(-2)(-4)$ so $k=-\frac{1}{2}$.

Therefore, the parabola has equation $y=-\frac{1}{2}(x-2)(x-4)$.

Next, we determine the coordinates of $D$, the vertex of the parabola.

Since the $x$-intercepts are 2 and 4 , then the $x$-coordinate of the vertex is the average of these, or 3.



The $y$-coordinate of $D$ can be obtained from the equation of the parabola; we obtain $y=-\frac{1}{2}(3-2)(3-4)=-\frac{1}{2}(1)(-1)=\frac{1}{2}$.

Thus, the coordinates of $D$ are $\left(3, \frac{1}{2}\right)$.

Lastly, we determine the area of $\triangle B D C$, whose vertices have coordinates $B(0,-4)$, $D\left(3, \frac{1}{2}\right)$, and $C(4,0)$.

Method 1 

We proceed be "completing the rectangle". That is, we draw the rectangle with horizontal sides along the lines $y=\frac{1}{2}$ and $y=-4$ and vertical sides along the lines $x=0$ and $x=4$. We label this rectangle as $B P Q R$.

![](https://cdn.mathpix.com/cropped/2023_12_21_d9b0e20a7552d07ba134g-1.jpg?height=271&width=266&top_left_y=710&top_left_x=1035)

The area of $\triangle B D C$ equals the area of the rectangle minus the areas of $\triangle B P D, \triangle D Q C$ and $\triangle C R B$.

Rectangle $B P Q R$ has height $4+\frac{1}{2}=\frac{9}{2}$ and width 4 .

$\triangle B P D$ has height $\frac{9}{2}$ and base 3 .

$\triangle D Q C$ has height $\frac{1}{2}$ and base 1.

$\triangle C R B$ has height 4 and base 4.

Therefore, the area of $\triangle B D C$ is $4\left(\frac{9}{2}\right)-\frac{1}{2}\left(\frac{9}{2}\right)(3)-\frac{1}{2}\left(\frac{1}{2}\right)(1)-\frac{1}{2}(4)(4)=18-\frac{27}{4}-\frac{1}{4}-8=3$.

Method 2

We determine the coordinates of $E$, the point where $B D$ crosses the $x$-axis.

![](https://cdn.mathpix.com/cropped/2023_12_21_d9b0e20a7552d07ba134g-1.jpg?height=277&width=309&top_left_y=1542&top_left_x=1011)

Once we have done this, then the area of $\triangle B D C$ equals the sum of the areas of $\triangle E C B$ and $\triangle E C D$.

Since $B$ has coordinates $(0,-4)$ and $D$ has coordinates $\left(3, \frac{1}{2}\right)$, then the slope of $B D$ is $\frac{\frac{1}{2}-(-4)}{3-0}=\frac{\frac{9}{2}}{3}=\frac{3}{2}$.

Since $B$ is on the $y$-axis, then the equation of the line through $B$ and $D$ is $y=\frac{3}{2} x-4$. To find the $x$-coordinate of $E$, we set $y=0$ to obtain $0=\frac{3}{2} x-4$ or $\frac{3}{2} x=4$ or $x=\frac{8}{3}$.

We think of $E C$ as the base of each of the two smaller triangles. Note that $E C=4-\frac{8}{3}=\frac{4}{3}$. Thus, the area of $\triangle E C D$ is $\frac{1}{2}\left(\frac{4}{3}\right)\left(\frac{1}{2}\right)=\frac{1}{3}$.

Also, the area of $\triangle E C B$ is $\frac{1}{2}\left(\frac{4}{3}\right)(4)=\frac{8}{3}$.

Therefore, the area of $\triangle B D C$ is $\frac{1}{3}+\frac{8}{3}=3$.

Final answer: 3