AIME 2000 I · 第 13 题

Solution

Let the intersection of the highways be at the origin $O$, and let the highways be the x and y axes. We consider the case where the truck moves in the positive x direction.

After going $x$ miles, $t=\frac{d}{r}=\frac{x}{50}$ hours has passed. If the truck leaves the highway it can travel for at most $t=\frac{1}{10}-\frac{x}{50}$ hours, or $d=rt=14t=1.4-\frac{7x}{25}$ miles. It can end up anywhere off the highway in a circle with this radius centered at $(x,0)$. All these circles are homothetic with respect to a center at $(5,0)$.

Now consider the circle at $(0,0)$. Draw a line tangent to it at $A$ and passing through $B (5,0)$. By the Pythagorean Theorem $AB^2+AO^2=OB^2 \Longrightarrow AB=\sqrt{OB^2-AO^2}=\sqrt{5^2-1.4^2}=\frac{24}{5}$. Then $\tan(\angle ABO)=\frac{OA}{AB}=\frac{7}{24}$, so the slope of line $AB$ is $\frac{-7}{24}$. Since it passes through $(5,0)$ its equation is $y=\frac{-7}{24}(x-5)$.

This line and the x and y axis bound the region the truck can go if it moves in the positive x direction. Similarly, the line $y=5-\frac{24}{7}x$ bounds the region the truck can go if it moves in positive y direction. The intersection of these two lines is $\left(\frac{35}{31},\frac{35}{31}\right)$. The bounded region in Quadrant I is made up of a square and two triangles. $A=x^2+x(5-x)=5x$. By symmetry, the regions in the other quadrants are the same, so the area of the whole region is $20x=\frac{700}{31}$ so the answer is $700+31=\boxed{731}$.

Intuition

Here's a bit of intuition behind the solution. If I wanted to travel to some point on the prairie, normally I would go the straight line distance. Right? However, here I would choose to travel on the highway the greatest distance, then drive the rest on the prairie. This is why we can easily use parametric equations to solve for the equations of lines that would yield those distances.

-WhatdoHumanitariansEat

Alternate finish

Once we get that the area consists of 8 right triangles with side lengths $1.4$, $4.8$, and $5$, then we can finish and find the area in an alternate way which uses no coordinates. Notice how we draw $45^{\circ}$ angles that connect opposite vertices such that the area can be expressed as just the sum of eight congruent triangles. So it suffices to find the area of one of them.

Here we have right triangle $\triangle OAB$ with $\angle OAB=90^{\circ}$, $OA=1.4$, $AB=4.8$, and $OB=5$. $C$ is the point on $AB$ such that $\angle COB=45^{\circ}$. We would like to find the area of $\triangle COB$, so we let $D$ be the foot of the altitude from $C$ to $OB$. We claim the following lemma.

CLAIM: $AD$ bisects $\angle OAB$

PROOF: We have $\angle OAC+\angle ODC=90^{\circ}$, so since $\angle OAC+\angle ODC=180^{\circ}$, then $OACD$ is cyclic. Additionally, in $\triangle OCD$, we have $\angle DOC=\angle DCO=45^{\circ}$, so $\angle OAD=\angle CAD=45^{\circ}$. This proves our lemma.

By the Angle Bisector Theorem, $\frac{OD}{BD}=\frac{AO}{AB}=\frac{7}{24}$. Therefore, $OD=\frac{7}{7+24}OB=\frac{35}{31}$, so $CD=OD=\frac{35}{31}$. Then the area of $\triangle OCB$ is $\frac{1}{2}bh=\frac{1}{2}\cdot 5\cdot \frac{35}{31}=\frac{175}{62}$. The area of the star figure consists of eight of these triangles, so its area is $\frac{175}{62}\cdot 8=\frac{700}{31}$. The answer is $\boxed{731}$.

~ethanzhang1001