MHB Construct a segment x satisfying the equation :1/x−2/a+3/(x+a)=0

[Albert1]

given a segment with length $a$
(1)construct a segment $x$ satisfying the equation :$\dfrac {1}{x}-\dfrac {2}{a}+
\dfrac {3}{x+a}=0$
(2)construct a segment $y$ with length $\sqrt {ax}$

 

[HallsofIvy]

Spoiler

For (1), first solve the equation for x: $$\frac{1}{x}- \frac{2}{a}+ \frac{3}{x+ a}= 0$$. Multiply both sides by ax(x+ a). [math]a(x+ a)- 2x(x+ a)+ 3a= -2x^2+ ax- 2ax+ a^2- 3a= -2x^- ax+ a^2= 0[/math]. which is equivalent to [math]2x^2+ ax- a^2= 0[/math], a quadratic equation. By the quadratic formula, [math]x= \frac{-a\pm \sqrt{a^2+ 4a^2}}{4}= \frac{-a\pm a\sqrt{5}}{4}[/math].

Basically, then the problem is to construct a segment of length [math]a\sqrt{5}[/math]. To do that, construct a segment of length 2a. At one end of that segment, construct the perpendicular to the segment and mark off length a on it. Draw the segment connecting the other two ends of those segments. This new segment is the hypotenuse of a right triangle with legs of length a and 2a so it has length \sqrt{a^2+ (2a)^2}= \sqrt{5a^2}= a\sqrt{5}.

 

[Albert1]

Spoiler

For (1), first solve the equation for x: $$\frac{1}{x}- \frac{2}{a}+ \frac{3}{x+ a}= 0$$. Multiply both sides by ax(x+ a). [math]a(x+ a)- 2x(x+ a)+ 3a= -2x^2+ ax- 2ax+ a^2- 3a= -2x^- ax+ a^2= 0[/math]. which is equivalent to [math]2x^2+ ax- a^2= 0[/math], a quadratic equation. By the quadratic formula, [math]x= \frac{-a\pm \sqrt{a^2+ 4a^2}}{4}= \frac{-a\pm a\sqrt{5}}{4}[/math].

Basically, then the problem is to construct a segment of length [math]a\sqrt{5}[/math]. To do that, construct a segment of length 2a. At one end of that segment, construct the perpendicular to the segment and mark off length a on it. Draw the segment connecting the other two ends of those segments. This new segment is the hypotenuse of a right triangle with legs of length a and 2a so it has length \sqrt{a^2+ (2a)^2}= \sqrt{5a^2}= a\sqrt{5}.

Spoiler

sorry ,calculation not correct (a typo)

 

[HallsofIvy]

Oh, Pooh! Yes, there is an error (calling it a "typo" is kind).

Spoiler

The original equation is \frac{1}{x}- \frac{2}{a}+ \frac{3}{x+ a}= 0.

Multiplying by ax(x+ a) gives a(x+ a)- 2x(x+ a)+ 3ax= 0, not what I had before. Then ax+ a^2- 2x^2- 2ax+ 3ax= -2x^2+ 2ax+ a^2= 0. That is a quadratic equation. By the quadratic formula \frac{-2a\pm\sqrt{4a^2+ 8a^2}}{-4}= \frac{-2a\pm\sqrt{12a^2}}{-4}= a\frac{1\pm\sqrt{3}}{2}

 

[Albert1]

Oh, Pooh! Yes, there is an error (calling it a "typo" is kind).

Spoiler

The original equation is \frac{1}{x}- \frac{2}{a}+ \frac{3}{x+ a}= 0.

Multiplying by ax(x+ a) gives a(x+ a)- 2x(x+ a)+ 3ax= 0, not what I had before. Then ax+ a^2- 2x^2- 2ax+ 3ax= -2x^2+ 2ax+ a^2= 0. That is a quadratic equation. By the quadratic formula \frac{-2a\pm\sqrt{4a^2+ 8a^2}}{-4}= \frac{-2a\pm\sqrt{12a^2}}{-4}= a\frac{1\pm\sqrt{3}}{2}

Spoiler

$x=a(\dfrac {1+\sqrt 3}{2})$ for $x=a(\dfrac {1-\sqrt 3}{2})$ is unreasonable
now how to construct segments with length $x=a(\dfrac {1+\sqrt 3}{2})$
and $y=\sqrt {ax}$

 

[lfdahl]

My attempt:

Spoiler

From
\[\frac{1}{x}-\frac{2}{a}+\frac{3}{x+a}=0\]
We get:
\[2x^2-2ax-a^2=0\Rightarrow x = \frac{1}{2}a+\frac{\sqrt{3}}{2}a = a\cos 30^{\circ} + a \sin 30^{\circ}\]

View attachment 7019

Construction of $x$: Divide a right angle (in the 1st quadrant) into three equal angles. This is done in three small steps with a compass. Let segment $a$ ($|OA|$) be the hypotenuse with an angle of 30 degrees to the horizontal line. From point $A$ draw a vertical line to meet the $x$-axis in point $B$. The adjacent catheter of triangle $OAB$ has the length: $\frac{\sqrt{3}}{2}a$, whereas the opposing catheter has the length $\frac{1}{2}a$. Now, add a half segment $a$ to the adjacent catheter (point $A’$). Our $x$ is then |OA’|.

View attachment 7020Construction of $ax$: Mark segment $a$ on the $y$-axis (from Origo, point $A$). Mark segment $x$ on the $x$-axis (from Origo, point $X$). Then mark the unit length (point $B$) on the $y$-axis. Now connect $B$ and $X$. Draw a line through point $A$ parallel to the line through $B$ and $X$. The line cuts the $x$-axis in point $C$ at a distance $ax$ from Origo. This is easily seen from the two triangles (with equal angles):
$\frac{|OX|}{|OB|} = \frac{|OC|}{|OA|} \rightarrow \frac{x}{1}=\frac{|OC|}{a} \rightarrow |OC| = ax$.View attachment 7021

Construction of $y = \sqrt{ax}$:
Mark segment $ax$ on the $x$-axis (point $A$). Then mark the point of unit length starting from Origo to the left (point $B$). Draw the half circle with diameter $|AB|$. The half circle and the $y$-axis cross in point ($Y$). We have two triangles with equal angles: OAY and OBY. Thus, we have the relation:

\[\frac{|Y|}{1} = \frac{|OA|}{|Y|}\rightarrow |Y|^2 = |OA| \rightarrow |Y| = \sqrt{|OA|}\rightarrow y = \sqrt{ax}\]

 

[Albert1]

My attempt:

Spoiler

From
\[\frac{1}{x}-\frac{2}{a}+\frac{3}{x+a}=0\]
We get:
\[2x^2-2ax-a^2=0\Rightarrow x = \frac{1}{2}a+\frac{\sqrt{3}}{2}a = a\cos 30^{\circ} + a \sin 30^{\circ}\]
Construction of $x$: Divide a right angle (in the 1st quadrant) into three equal angles. This is done in three small steps with a compass. Let segment $a$ ($|OA|$) be the hypotenuse with an angle of 30 degrees to the horizontal line. From point $A$ draw a vertical line to meet the $x$-axis in point $B$. The adjacent catheter of triangle $OAB$ has the length: $\frac{\sqrt{3}}{2}a$, whereas the opposing catheter has the length $\frac{1}{2}a$. Now, add a half segment $a$ to the adjacent catheter (point $A’$). Our $x$ is then |OA’|.

Construction of $ax$: Mark segment $a$ on the $y$-axis (from Origo, point $A$). Mark segment $x$ on the $x$-axis (from Origo, point $X$). Then mark the unit length (point $B$) on the $y$-axis. Now connect $B$ and $X$. Draw a line through point $A$ parallel to the line through $B$ and $X$. The line cuts the $x$-axis in point $C$ at a distance $ax$ from Origo. This is easily seen from the two triangles (with equal angles):
$\frac{|OX|}{|OB|} = \frac{|OC|}{|OA|} \rightarrow \frac{x}{1}=\frac{|OC|}{a} \rightarrow |OC| = ax$.

Construction of $y = \sqrt{ax}$:
Mark segment $ax$ on the $x$-axis (point $A$). Then mark the point of unit length starting from Origo to the left (point $B$). Draw the half circle with diameter $|AB|$. The half circle and the $y$-axis cross in point ($Y$). We have two triangles with equal angles: OAY and OBY. Thus, we have the relation:

\[\frac{|Y|}{1} = \frac{|OA|}{|Y|}\rightarrow |Y|^2 = |OA| \rightarrow |Y| = \sqrt{|OA|}\rightarrow y = \sqrt{ax}\]

lfdahl,thanks for participation. well done!

 

[HallsofIvy]

Just in general: The "constructible numbers" are precisely those numbers that are "algebraic of order a power of 2". That is, those numbers that satisfy a polynomial equation with integer coefficients with degree a power of 2.