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Can a triangle be constructed by right edge and compass when its base, median and the sum of other sides are known?

I found this problem in the book "An introduction to the modern geometry of the triangles and the circles" by Nathan Altshiller Court. This is my effort:

Case 1: The measure of median is greater than half base:

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We have base AB and median EC; first we do following calculations:

$BE^2=CE^2-BC^2$

$AE^2=BE^2+AB^2$

$AE+EB=\sqrt{BE^2+AB^2}+\sqrt{CE^2-BC^2}$

therefore the sum of two other sides given must be competent with this calculation, it will not be necessary to know it. Now for construction of triangle we have to draw an ellipse, its intersection with a circle center at C and radius equal to median will be the third vertex. For drawing the ellipse we may use ancient method; using a rope length equal to $l=AB + BE+AE$ and a pen which make a triangle with A and B. Moving this pen give us the ellipse. I have no idea how to do this using a compass.

Case 2- The measure of median is less than half base:

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First we draw ellipse using three center arc which is called ellipse like arc. First we divide AB in three equal segments. Construct an equilateral triangle CDF on middle part CD. Draw two circle on centers E and D, with radii $EB=DA$, then draw third circle center on F , radius $2 EB=\frac 23 AB$. This arc is tangent on previous two arcs at points I and K and the arc AIKB is approximately a part of an ellipse passing points A and B. Now we draw a circle center on C and radius equal to the measure of median. The intersection of this circle and the ellipse will be the third vertex of triangle. As can be seen in figure , the third vertex is intersection of median circle with circle center of F and $r=\frac 23 AB $if median $m<\frac 13 BC$ (point L in figure) and it is with circle center on E and radius $r=\frac 13 AB$ if median $BC>m>\frac 12 BC$ (point N in the figure).

Again the sum of two other sides, if given, is not used.

Question: How about if the the sum of measures of other sides is given? any idea for constructing the triangle with straight edge and compass?

Note: the original problem was the difference of two other sides is given. In this case the third vertex can be the intersection of the circles g and h with a hyperbola. I tried to solve a simpler form; sum instead of difference.

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  • $\begingroup$ The circle with center in A and radius AB intersect the circle with center in M and radius the median in C; if the intersection is not empty. $\endgroup$
    – lib
    Aug 25, 2023 at 14:26

1 Answer 1

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The problem can be formulated so:

Let $M$ be the midpoint of the side $AB$ in the triangle $ABC$. Let $CX$ be perpendicular drawn from $C$ to $AB$. Is it possible to construct $x=MX$ with right edge and compass if $AB$, $CM$ and $AC+BC$ are known?

Let $CM=m$, $AB=2a$, $AC+BC=2l$.

Then the last equation can be written (see figure): $$ \sqrt{(a+x)^2+(m^2-x^2)}+\sqrt{(a-x)^2+(m^2-x^2)}=2l. $$ Squaring the equation one obtains: $$\begin{align} &2a^2+2m^2+2\sqrt{(a^2+2ax+m^2)(a^2-2ax+m^2)}=4l^2\\ \implies&(2l^2-a^2-m^2)^2=(a^2+2ax+m^2)(a^2-2ax+m^2)=(a^2+m^2)^2-4a^2x^2\\ \iff&4a^2x^2=(a^2+m^2)^2-(2l^2-a^2-m^2)^2=2l^2(2a^2+2m^2-2l^2)\\ \iff& a|x|=l\sqrt{a^2+m^2-l^2}\tag1 \end{align}$$

Given $a,m,l$ the equation $(1)$ obviously can be solved for $x$ with compass and right edge, provided that $l>a$ and $\sqrt{l^2-a^2}\le m<l$.

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    $\begingroup$ Of course $x=MX$, not $CX$. $\endgroup$ Aug 27, 2023 at 20:03
  • $\begingroup$ Of course. Thank you for pointing this out! $\endgroup$
    – user
    Aug 27, 2023 at 20:04
  • $\begingroup$ $user, Perfect. thank you. $\endgroup$
    – sirous
    Aug 27, 2023 at 21:25

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