The field shown has been planted uniformly with wheat. At harvest, the wheat at any point in the field is brought to the nearest point on the field's perimeter. The fraction of the crop that is brought to the longest side is
(A) $\frac{1}{3}$
(B) $\frac{5}{12}$
(C) $\frac{1}{2}$
(D) $\frac{2}{5}$
(E) $\frac{4}{9}$

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-089.jpg?height=217&width=442&top_left_y=1057&top_left_x=1297)
We first note that the given quadrilateral is a trapezoid, because $60^{\circ}+120^{\circ}=180^{\circ}$, and so the top and bottom sides are parallel.

We need to determine the total area of the trapezoid and then what fraction of that area is closest to the longest side.

Area of trapezoid

Label the trapezoid as $A B C D$ and drop perpendiculars from $B$ and $C$ to $P$ and $Q$ on $A D$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-170.jpg?height=252&width=458&top_left_y=1744&top_left_x=888)

Since $\triangle A B P$ is right-angled at $P$ and $\angle B A P=60^{\circ}$, then $A P=100 \cos \left(60^{\circ}\right)=100\left(\frac{1}{2}\right)=50 \mathrm{~m}$ and $B P=100 \sin \left(60^{\circ}\right)=100\left(\frac{\sqrt{3}}{2}\right)=50 \sqrt{3} \mathrm{~m}$. (We could have used the ratios in a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle to do these calculations.)

By symmetry, $Q D=50 \mathrm{~m}$ as well.

Also, since $B C$ is parallel to $P Q$, and $B P$ and $C Q$ are perpendicular to $P Q$, then $B P Q C$ is a rectangle, so $P Q=B C=100 \mathrm{~m}$.

Thus, the area of trapezoid $A B C D$ is $\frac{1}{2}(B C+A D)(B P)=\frac{1}{2}(100+(50+100+50))(50 \sqrt{3})$ or $7500 \sqrt{3}$ square metres.

Determination of region closest to $A D$

Next, we need to determine what region of the trapezoid is closest to side $A D$.

To be closest to side $A D$, a point inside the trapezoid must be closer to $A D$ than to each of $B C, A B$, and $D C$.

For a point in the trapezoid to be closer to $A D$ than to $B C$, it must be below the "half-way mark", which is the midsegment $M N$.

Thus, such a point must be below the parallel line that is $\frac{1}{2}(50 \sqrt{3})=25 \sqrt{3} \mathrm{~m}$ above $A D$.

For a point in the trapezoid to be closer to $A D$ than to $A B$, it must be below the angle bisector of $\angle B A D$. (See the end of this solution for a justification of this.)

Similarly, for a point in the trapezoid to be closer to $A D$ than to $D C$, it must be below the angle bisector of $\angle C D A$.

Define points $X$ and $Y$ to be the points of intersection between the angle bisectors of $\angle B A D$ and $\angle C D A$, respectively, with the midsegment $M N$. We will confirm later in the solution that the angle bisectors intersect above $M N$, not below $M N$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-171.jpg?height=236&width=458&top_left_y=817&top_left_x=888)

Area of trapezoid $A X Y D$

Lastly, we need to determine the area of trapezoid $A X Y D$.

Note that $\angle X A D=\angle Y D A=\frac{1}{2}\left(60^{\circ}\right)=30^{\circ}$.

Drop perpendiculars from $X$ and $Y$ to $G$ and $H$, respectively, on $A D$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-171.jpg?height=244&width=453&top_left_y=1363&top_left_x=890)

We know that $A D=200 \mathrm{~m}$ and $X G=Y H=25 \sqrt{3} \mathrm{~m}$.

Since each of $\triangle A X G$ and $\triangle D Y H$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle, then

$$
A G=D H=\sqrt{3} X G=\sqrt{3}(25 \sqrt{3})=75
$$

This tells us that the angle bisectors must intersect above $M N$, since $A G+H D=150$ and $A D=200$, so $A G+H D<A D$.

Since $X G H Y$ is a rectangle (by similar reasoning as for $B P Q C$ ), then

$$
X Y=G H=A D-A G-D H=200-75-75=50
$$

Therefore, the area of trapezoid $A X Y D$ is $\frac{1}{2}(A D+X Y)(X G)=\frac{1}{2}(50+200)(25 \sqrt{3})$ or $3125 \sqrt{3}$ square metres.

This tells us that the fraction of the crop that is brought to $A D$ is $\frac{3125 \sqrt{3}}{7500 \sqrt{3}}=\frac{25}{60}=\frac{5}{12}$.

Property of angle bisectors

We must still justify the fact about angle bisectors above.

Consider $\angle S T U$ and point $V$ on its angle bisector.

Drop perpendiculars from $V$ to $E$ and $F$ on $S T$ and $U T$, respectively.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-172.jpg?height=272&width=352&top_left_y=154&top_left_x=949)

Now $\triangle V E T$ is congruent to $\triangle V F T$ since each is right-angled, the two have equal angles at $T$, and the two share a common hypotenuse $T V$.

This tells us that $E V=F V$; that is, $V$ is the same distance from $S T$ and from $U T$.

This also tells us that any point on the angle bisector will be equidistant from $S T$ and $U T$.

We can also deduce from this that any point below the angle bisector will be closer to $U T$, as moving from a point on the angle bisector to such a point moves closer to $U T$ and further from $S T$.

ANswer: (B)