In the adjoining figure, the triangle ABC is a right triangle with ∠ BCA=90°. Median CM is perpendicular to median BN, and side BC=s. The length of BN is

1982 AMC 12 Problem 21

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21.

In the adjoining figure, the triangle

ABC

is a right triangle with

\angle BCA=90^\circ.

Median

CM

is perpendicular to median

BN,

and side

BC=s.

The length of

BN

s\sqrt2

\frac32s\sqrt2

2s\sqrt2

\frac{s\sqrt5}{2}

\frac{s\sqrt6}{2}

Place

C=(0,0),

B=(0,s),

and

A=(t,0)

Use a dot product for the directions of the two medians

Set

C=(0,0),

B=(0,s),

and

A=(t,0).

Then

M=(\frac{t}{2},\frac{s}{2})

and

N=(\frac{t}{2},0).

Perpendicularity gives

(\frac{t}{2},\frac{s}{2})\cdot(\frac{t}{2},-s)=0,

hence

t^2=2s^2.

Therefore

\begin{aligned} BN&=\sqrt{(\frac{t}{2})^2+s^2}\\ &=\sqrt{\frac{s^2}{2}+s^2}\\ &=\frac{s\sqrt6}{2}. \end{aligned}

Therefore, the correct answer is E.