SAT1000 - P897

Geometry Level 4

In the rectangular coordinate plane, the Hamilton distance from point $P_1(x_1,y_1)$ to $P_2(x_2,y_2)$ is defined as: $d(P_1,P_2)=|x_1-x_2|+|y_1-y_2|$

Which of the following three statements are true ?

If point $C$ is on segment $AB$ , then $d(A,C)+d(C,B)=d(A,B)$ .
In $\triangle ABC$ , if $\angle C=90 \degree$ , then $d^2(A,C)+d^2(C,B)=d^2(A,B)$ $(d^2(A,B)=(d(A,B))^2)$ .
In $\triangle ABC$ , $d(A,C)+d(C,B)>d(A,B)$ .

How to submit:

Let $p_1, p_2,\cdots,p_n$ be the boolean value of statement $1,2,\cdots,n$ , if statement $k$ is true, $p_k=1$ , else $p_k=0$ . Then submit $\displaystyle \sum_{k=1}^n p_k \cdot 2^{k-1}$ .

Have a look at my problem set: SAT 1000 problems

The answer is 1.

2 solutions

Ved Pradhan
Jun 30, 2020

This is all about visualizing Taxicab Geometry!

The first statement is obviously true. Even if you're measuring the object in blocks, it makes sense that going from $A$ to $C$ and then $C$ to $B$ should provide the same result as going from $A$ to $B$ directly. Just think as if you want to go to your friend's house, and then your grandparents' house. Shouldn't that provide the same result as going to your grandparents house? Note that it actually doesn't matter if the house are not collinear! (We'll get back to this in point three).

If this is confusing, think of it as vector addition using $x$ and $y$ components.

The second statement refers to the Pythagorean theorem. However, this statement must be false. Why? Taking the square root of the hypotenuse gives us what is known as the distance formula. However, in Taxicab geometry, distance itself takes a different formula! So clearly, this statement must be false.

Last but not least, let's go to the third statement. Remember that analogy I gave about the two houses? I said they didn't have too be collinear! You might think, "oh, but what if the first house is farther out than the second house? Then the distance is larger!" Unfortunately, that logic doesn't make sense, because here, going "backwards" counts as negative length. Just think about it as if you're doing vector addition with $x$ and $y$ components.

With that, let's get our answer:

$1 \times 2^{0} + 0 \times 2^{1} + 0 \times 2^{2} = \boxed{1}$

Alapan Das
Jul 1, 2020

$d(P,Q)$ is simply the operation of first moving from $P$ in horizontal direction and then vertically to $Q$ .

Meaning, If we form a right triangle $\triangle{P\psi Q}$ such that $P\psi$ is horizontal (say) and $\psi Q$ is vertical.

Then, $d(P,Q)=\overline{p\psi}+ \overline{\psi Q}$ .....(*)

This directly implies that 1) is true.

And 2) isn't true. It's true for different metric $d(P,Q)=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$ .

3) is also not true. (*) directly implies that the inequality of triangle is replaced by equality in some cases in spite of making a perfect triangle. This should be

$d(A,C)+d(C,B)≥d(A,B)$ .

The equality holds when $C$ is inside the rectangle formed by $A\psi_{A,B} B \psi'_{A,B}$ .

SAT1000 - P897

The answer is 1.

2 solutions

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