SAT1000 - P877

Geometry Level pending

Here's the definition of harmonically bisect : Given that $A_1, A_2, A_3, A_4$ are four distinct points on the coordinate plane, if $\overrightarrow{A_1A_3}=\lambda \overrightarrow{A_1A_2}$ , $\overrightarrow{A_1A_4}=\mu \overrightarrow{A_1A_2}$ , $\dfrac{1}{\lambda}+\dfrac{1}{\mu}=2$ , then $A_3, A_4$ harmonically bisect $A_1, A_2$ .

Given that $C(c,0), D(d,0)\ (c,d \in \mathbb R)$ harmonically bisect $A(0,0), B(1,0)$ , which choice is true ?

$A.\ \textup{C could be the midpoint of AB.}$

$B.\ \textup{D could be the midpoint of AB.}$

$C.\ \textup{C, D could be both on segment AB.}$

$D.\ \textup{C, D can't be both on the extension line of AB.}$

Have a look at my problem set: SAT 1000 problems

D

B

C

A

1 solution

A Former Brilliant Member
Jun 22, 2020

From the given definition we get $\dfrac {1}{c}+\dfrac {1}{d}=2$ .

If $C$ be the mid point of $\overline {AB}$ , then $D$ is at infinity. Similarly, if $D$ be the mid point of $\overline {AB}$ , then $C$ is at infinity.

Both of $c$ and $d$ can not lie between $0$ and $1$ simultaneously, since then $\dfrac {1}{c}+\dfrac {1}{d}$ will be greater than $2$ . So points $C$ and $D$ both can not lie within the segment $\overline {AB}$ .

So if any one of $c$ and $d$ be greater than $1$ , the other has to be less than $1$ . Hence, D is the correct option .

SAT1000 - P877

1 solution

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