ARML Power Round - 2

Geometry Level 4

Let the taxicab distance between two points be defined by:

D t = ∣ x 2 − x 1 ∣ + ∣ y 2 − y 1 ∣ D_t = |x_2-x_1|+|y_2-y_1|

Let a taxicab circle be the set of all point that are equidistant (in taxicab distance) from a fixed point, referred to as the center of the taxicab circle.

A taxicab circle is what geometric shape?

This problem is not original, it comes from the ARML 2016 Power Round

Hexagon Triangle None of these Pentagon Circle Square

Alex G by Alex G , 22, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Hung Woei Neoh
Jun 5, 2016

This question is best explained with an example.

Now, let us define a taxicab circle with center ( x 1 , y 1 ) = ( 0 , 0 ) (x_1,y_1) =(0,0) and D t = 4 D_t=4 . The equation of the taxicab circle is then:

4 = ∣ x 2 ∣ + ∣ y 2 ∣ 4=|x_2|+|y_2|

We know that:

∣ x 2 ∣ = { x 2 x 2 ≥ 0 − x 2 x 2 < 0 ∣ y 2 ∣ = { y 2 y 2 ≥ 0 − y 2 y 2 < 0 |x_2| = \begin{cases} x_2&\quad x_2 \geq 0\\ -x_2&\quad x_2 < 0\end{cases}\\ |y_2| = \begin{cases} y_2&\quad y_2 \geq 0\\ -y_2&\quad y_2 < 0\end{cases}

Therefore, the equation 4 = ∣ x 2 ∣ + ∣ y 2 ∣ 4=|x_2|+|y_2| can be written as:

{ 4 = x 2 + y 2 x 2 ≥ 0 , y 2 ≥ 0 4 = x 2 − y 2 x 2 ≥ 0 , y 2 < 0 4 = − x 2 + y 2 x 2 < 0 , y 2 ≥ 0 4 = − x 2 − y 2 x 2 < 0 , y 2 < 0 \begin{cases} 4=x_2+y_2&\quad x_2 \geq 0, y_2 \geq 0\\ 4=x_2-y_2&\quad x_2 \geq 0, y_2 < 0\\ 4=-x_2+y_2&\quad x_2 <0, y_2 \geq 0\\ 4=-x_2-y_2&\quad x_2 < 0, y_2 <0 \end{cases}

Draw out these 4 4 lines, and you will get this shape:

You can go ahead and prove that all 4 4 sides are of the same length, and the vertices all have a right angle (connecting lines are perpendicular)

Therefore, the taxicab circle is actually a Square \boxed{\text{Square}}

Note: You can try this with any other center ( x 1 , y 1 ) (x_1,y_1) and any other positive real value of D t D_t . You will still get a square with a different position or size

0 Helpful 0 Interesting 0 Brilliant 0 Confused

You can also add that when x 2 , y 2 ≠ 0 x_{2}, y_{2} \ne 0 it is just the same figure with shifted origin.

A Former Brilliant Member - 5 years ago

Log in to reply

Thanks for the suggestion, I've added it in

Hung Woei Neoh - 5 years ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...