I'm not sure if my solution to this Olympiad Geometry question is valid.

Let $\square ABCD$ be a square with side length $1$. Let $E$, $P$, $F$ be the midpoints of $AD$, $CE$, $BP$, respectively. What is the area of $\triangle BFD$?

To solve this I used coordinate geometry but I'm not sure if it works.

I dropped a perpendicular from point P to side CD and called the point X. Then, since the area of triangle DPB is exactly double the desired area, I assumed point D to be (0, 0) and found the side length XP using the Pythagorean theorem. Since I have the three coordinates for triangle DPB, I used shoelace formula to find the area and divided by 2 to get {(19)^(1/2) - 1} / 8

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    $\begingroup$ Welcome to MSE. Is this contest on-going? $\endgroup$ – Rhys Hughes Jan 21 at 21:15
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    $\begingroup$ As this is for "contest math", please include the source of this question. In what publication (title of publication, author, date of publication), contest (provider of the contest and the name and date of the contest), etc., did you encounter this question? $\endgroup$ – jordan_glen Jan 21 at 21:16
  • $\begingroup$ I came across this question on a long list of contest problems but unfortunately, the source of the question is not provided. To be honest, I'm not sure if the question came up on an actual contest. Does that mean I should remove the tag? $\endgroup$ – gm_dalkomm Jan 21 at 21:20
  • $\begingroup$ I had the result $\text{Area}=\frac{1}{16}$ $\endgroup$ – Dr. Mathva Jan 21 at 21:24
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    $\begingroup$ @Blue apologies, my edit still in review. I did not notice the timing of your last edit nor that my edit suggestion is in review. Hence I deleted my comment as soon as I realized that. $\endgroup$ – jordan_glen Jan 21 at 21:28

Using Euclidean geometry you get straight to the result.

Distance of $P$ from $DC$ is $\frac{1}{4}$ and distance of $P$ from $BC$ is $\frac{1}{2}$, by Thales theorem. So you obtain the area of $DPB$ by successive subtractions from the area of $ABCD$, then divide by two to get the desired area: \begin{eqnarray} \mathcal A_{BFD}&=& \frac{1}{2}\cdot\left(\mathcal A_{ABCD}-\mathcal A_{ABD}- \mathcal A_{BCP}-\mathcal A_{CDP}\right)=\\ &=& \frac{1}{2}\cdot\left(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}\right)=\\ &=&\frac{1}{16}. \end{eqnarray}



The equation of the line $\overleftrightarrow{BD}$ is $x+y-1=0$. The coordinates of the point $F$ are $\left(\dfrac 34, \dfrac 38 \right)$. So the distance from the point $F$ to the line $\overleftrightarrow{BD}$ is $$h = \dfrac{\left|\dfrac 34 + \dfrac 38 - 1 \right|}{\sqrt{1^2+1^2}} = \dfrac{1}{8 \sqrt 2}$$

The area of $\triangle BDF$ is therefore $$\dfrac 12 \cdot BD \cdot h \ = \dfrac 12 \cdot \sqrt 2 \cdot \dfrac{1}{8 \sqrt 2} = \dfrac{1}{16}$$

An easier-to-remember way to write @Rosenberg's answer is

$$BDF = \dfrac 12 \cdot \left\| \begin{array}{c} 1 & 1 & 1 \\ 1 & 0 & \dfrac 34 \\ 0 & 1 & \dfrac 38 \end{array} \right\| = \dfrac{1}{16}$$

where $\| \cdot \|$ indicates the absolute value of the determinant.


The distance from the point $(h,k)$ to the line $ax+by-c=0$ is $\dfrac{|ah+bk-c|}{\sqrt{a^2+b^2}}$

The area of the triangle whose vertices have coordinates $(a_1, a_2), (b_1, b_2),(c_1, c_2)$ is $$\dfrac 12 \cdot \left\| \begin{array}{c} 1 & 1 & 1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{array} \right\|$$


Let $A(0,0)$, $B(0,1),$ $C(1,1)$ and $D(1.0).$

Thus, $E(0.5,0)$, $P(0.75,05)$ and $F(0.375,0.75).$

Now, use $$S_{\Delta BFD}=\frac{1}{2}\left|x_B(y_F-y_D)+x_F(y_D-y_B)+x_D(y_B-y_F)\right|.$$ I got $$S_{\Delta BDF}=\frac{1}{16}.$$


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