Divide a line segment in the ratio $\sqrt{2}:\sqrt{3}.$

Question

"Divide a line segment in the ratio $\sqrt{2}:\sqrt{3}.$"

I have got this problem in a book, but I have no idea how to solve it.

Any help will be appreciated.

Answer 1

To draw a line segment with ratio $a:b$ notice the total length of the line segment will be $n * (a + b),$ where $n$ is an arbitrary factor. Set $$L = n*(a + b)$$ where $L$ is the length of the line segment to find $n.$

Does that give you enough to complete the solution?

Answer 2

I assume you need to construct said division.

Let $AB$ be the line segment to be divided. Construct the semilines $r$ and $s$ perpendicular to $AB$ through $A$ and $B$, respectively such that they are both in the same "side" of $AB$.

Draw a circle of radius $AB$ from $A$. It intersects $r$ in $C$. Draw a circle of radius $BC$ from $B$. It intersects $s$ in $D$. Now $BD = \sqrt{2}AB$. Draw a paralel line to $AB$ from $D$. It intersects $r$ in $E$. Now $BE = \sqrt{3}AB$. Draw a circle of radius $BE$ from $D$. It intersects $s$ in $F$. Draw the line $AF$ and a paralel to $AF$ from $D$. It intersects $AB$ in $O$.

Now $AO:OB = \sqrt{3}:\sqrt{2}$.

Answer 3

The key is to remember that in any triangle, if you bisect an angle it cuts the opposite side into the same proportion as the sides of the angle. (Source: http://hotmath.com/hotmath_help/topics/triangle-angle-bisector-theorem.html). So to cut a segment into the ratio $\sqrt{2}:\sqrt{3}$, it is enough to bisect a triangle whose sides have that ratio.

Start with the given segment $AB$. Using that as a base, construct an isosceles right triangle (I assume you know how to construct perpendiculars.) Then the hypotenuse $AC$ of that triangle has length equal to $AB\sqrt{2}$.

Now construct right triangle $ABD$ as shown below. It is easy to see that $BD=AB\sqrt{2}$ and $AD=AB\sqrt{3}$. So if you bisect $\angle ADB$, it will cut $AB$ into the desired ratio.

Answer 4

Alternatively, let $AB$ be the line segment you want to divide with a point $P$ such that $\frac{|AP|}{|PB|}=\frac{\sqrt{2}}{\sqrt{3}}$. Let $C$ be on the extension of $AB$ such that $B$ lies between $A$ and $C$ and that $|BC|=6\,|AB|$. Draw the circle $\Gamma$ with diameter $AC$. The line perpendicular to $AC$ at $B$ meets $\Gamma$ at $D$ and $E$. Let $Q$ be the point on $DE$ such that $|DQ|=3\,|AB|$. Then, the circle centered at $B$ with radius $BQ$ meets the interior of $AB$ at the required point $P$.

Answer 5

let AB be the given line Segment. Draw a line L1 from A which makes an angle 45 degrees, and line L2 from B which makes an angle 60 with AB. Let L1, L2 meet at C. Now ABC is a triangle with angle A= 45, angle B = 60.
Let the angular bisector of C (makes an angle 37.5 with either of AC,BC) meets AB at D.
We know that $$AD/DB = AC/BC ..........(1)$$
we also know that
$$AC/sin(97.5) = CD/sin45$$$$ BC/sin(82.5) = CD/sin60$$
from the triangles CAD, CBD respectively
combining the above two equations
$$AC/BC = sin60/sin45 = (\sqrt{3}/2 )/(1/\sqrt{2}) = \sqrt{3}/\sqrt{2}$$ from (1)
$$AD/DB = \sqrt{3}/\sqrt{2}$$
Note that sin97.5 = sin 82.5
our line is divided by the point D in the required ratio.

Answer 6

For ease of exposition, let's assume $|AB|=1$. Now draw a perpendicular to $AB$ at $B$ and let $C$ and $D$ be points on the perpendicular with $C$ between $B$ and $D$ and $|BC|=|CD|=1$. Next draw circles of radius $|BD|=2$ centered at $B$ and $D$, and let $E$ be a point of intersection of those two circles. Note that $\triangle BDE$ is equilateral, with sides of length $2$, and $C$ is the midpoint of the side $BD$, so $|CE|=\sqrt3$.

Now draw a circle of radius $|CE|=\sqrt3$ centered at $C$, and then extend $AC$ until it intersects that circle at a point $F$, with $C$ between $A$ and $F$. Note that $|AC|=\sqrt2$ and $|CF|=\sqrt3$.

Finally, construct the line through $C$ parallel to $BF$. (Constructing parallel lines is a standard straightedge-and-compass procedure.) This line intersects $AB$ at a point $P$, and we have

$$|AP|:|PB|=|AC|:|CF|=\sqrt2:\sqrt3$$

Added later: It occurs to me it might be worth counting the total number of distinct straightedge-and-compass steps involved. Assuming we start with the extended line $AB$ (not just the line segment), constructing the perpendicular at $B$ can be done with $3$ circles and $1$ line; if you do it carefully (by first drawing the circle centered at $B$ of radius $|AB|$), you get the point $C$. It then takes $1$ more circle to get $D$ and $2$ circles to get $E$. To get $F$ takes a circle and a line. Finally, constructing the parallel to $FB$ through $C$ can be done with $3$ circles and $2$ lines: draw the line $FB$, draw the circle centered at $F$ of radius $|CF|$, intersecting $FB$ at $Q$, followed by circles centered at $C$ and $Q$ of radius $|CF|=|QF|$, which intersect at $R$, and finally the line $CR$, which is parallel to $FB$ (and intersects $AB$ at the desired point $P$). Thus the entire construction involves a total of $10$ compass steps and $4$ straightedge steps. (If all you have to start with are the two points $A$ and $B$, one more straightedge step, to construct the line $AB$, is needed.)

I'd be interested to know if any of the other answers' constructions get by with fewer steps.

Answer 7

Picture first. Wordy description to follow.

We seek a point $C$ that divides $\overline{AB}$ into segments in proportion $\sqrt{2}:\sqrt{3}$.

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Construct the line perpendicular to $\overleftrightarrow{AB}$ at $A$. (Note: The perpendicular at any point would work, but this gives me a named point to work with.)

Using any unit you like, mark-off eight congruent segments on the perpendicular. Let $C^{\prime\prime}$, $A^\prime$, and $B^\prime$ be the second, sixth, and eighth constructed endpoints of those segments, as shown in the diagram. Construct semicircles with diameter $\overline{AA^\prime}$ and $\overline{BB^\prime}$. (Note that the centers of these semicircles are conveniently among the constructed endpoints; that's actually why we constructed so many of them!)

Construct the perpendicular to $\overleftrightarrow{AA^\prime}$ at $C^{\prime\prime}$, and let $A^{\prime\prime}$ and $B^{\prime\prime}$ be the points where this line meets the semicircles. $\triangle AA^\prime A^{\prime\prime}$ is therefore a right triangle from which we derive the proportion $$\frac{|\overline{AC^{\prime\prime}}|}{|\overline{A^{\prime\prime} C^{\prime\prime}}|} = \frac{|\overline{A^{\prime\prime} C^{\prime\prime}}|}{|\overline{A^\prime C^{\prime\prime}}|} \;\to\; |\overline{A^{\prime\prime} C^{\prime\prime}}|^2 = |\overline{A C^{\prime\prime}}||\overline{A^{\prime}C^{\prime\prime}}| = 2 |\overline{A C^{\prime\prime}}|^2 \;\;\to\;\; |\overline{A^{\prime\prime}C^{\prime\prime}}| = \sqrt{2}\,|\overline{A C^{\prime\prime}}|$$ Likewise, $|\overline{B^{\prime\prime}C^{\prime\prime}}| = \sqrt{3}\,|\overline{AC^{\prime\prime}}|$, so that $C^{\prime\prime}$ divides $|\overline{A^{\prime\prime}B^{\prime\prime}}|$ into lengths in proportion $\sqrt{2} : \sqrt{3}$.

Consequently, if $X$ is the intersection of $\overleftrightarrow{AA^{\prime\prime}}$ and $\overleftrightarrow{BB^{\prime\prime}}$, then line $\ell := \overleftrightarrow{XC^{\prime\prime}}$ meets $\overline{AB}$ at the desired division point $C$. (If it happens that $\overleftrightarrow{AA^{\prime\prime}}$ and $\overleftrightarrow{BB^{\prime\prime}}$ are parallel, then take $\ell$ to be the line, through $C^{\prime\prime}$, parallel to them.) $\square$

Notes:

• We get the same $C$ if we construct each semicircle on the "other side" of the perpendicular line.

• If we construct both semicircles on the "same side" of the perpendicular line, then we obtain a point $C$ external to $\overline{AB}$ yet with $|\overline{AC}|:|\overline{BC}| = \sqrt{2}:\sqrt{3}$. (We get the same external $C$, regardless of which side is the "same side".)

• This strategy can be used to subdivide a segment into any proportion of the form $\sqrt{m} : \sqrt{n}$.

Answer 8

Draw an isosceles rectangle triangle. Carry the hypothenuse on a side. If the hypothenuse is $\sqrt2$, the new segment is $\sqrt3$.

By Thales' theorem you can divide any segment in this ratio.