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Is it possible to draw a triangle, if the length of its medians $(m_1, m_2, m_3)$ are given only?

Someone asked me this question, but I can not see it. Is it really possible?

UPDATE

Apart from the algebraic solution given by JimmyK4542, can anyone give me a direct construction? I mean, it should sound like:

Draw a line segment sufficiently long. Cut the length of $m_1$ from it. Then $\ldots$

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    $\begingroup$ How to construct a triangle given its three medians. $\endgroup$ – user856 Jul 23 '14 at 4:03
  • $\begingroup$ @Rahul I got it! Thank you. $\endgroup$ – hola Jul 23 '14 at 4:15
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From JimmyK4542's formulas it follows that $a={2\over3}\sqrt{2s^2-m_a^2}\>$, where $s:=\sqrt{m_b^2+m_c^2}$. From this one derives the following construction of ${3\over2}a$:

enter image description here

Construct $s$ as hypotenuse of a right triangle with legs $m_b$, $m_c$; then draw a square with side $s$ and find $\sqrt{2}\>s$ as length of the diagonal $d$. Draw a Thales semicircle with diameter $d$; then construct a right triangle with hypotenuse $d$ and one leg $m_a$. The other leg then is ${3\over2}a$. The rest should be easy.

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  • $\begingroup$ Can you clarify that construction? I can see how to get the line of length $3a/2$ as the common tangent of two circles, but since that's not indicated in the picture it's not clear to me that's what you intended. $\endgroup$ – Semiclassical Jul 23 '14 at 12:20
  • $\begingroup$ That helps. What threw me is that it wasn't clear to me that the isoceles triangle with $d$ as its diagonal is a right triangle. $\endgroup$ – Semiclassical Jul 23 '14 at 13:43
  • $\begingroup$ @Semiclassical: Hope its clear now. $\endgroup$ – Christian Blatter Jul 23 '14 at 13:43
  • $\begingroup$ Can you complete the construction with steps? $\endgroup$ – user142795 Sep 17 '14 at 10:38
  • $\begingroup$ @user142795: The steps are described below the figure. $\endgroup$ – Christian Blatter Sep 17 '14 at 10:59
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The formulas for the lengths of the medians of a triangle given the sidelengths are:

$m_a^2 = \dfrac{2b^2+2c^2-a^2}{4}$

$m_b^2 = \dfrac{2c^2+2a^2-b^2}{4}$

$m_c^2 = \dfrac{2a^2+2b^2-c^2}{4}$

Solving for $a,b,c$ in terms of $m_a,m_b,m_c$ gives:

$a^2 = \dfrac{8m_b^2+8m_c^2-4m_a^2}{9}$

$b^2 = \dfrac{8m_c^2+8m_a^2-4m_b^2}{9}$

$c^2 = \dfrac{8m_a^2+8m_b^2-4m_c^2}{9}$

This gives you the lengths of the three sides of the triangle.

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  • $\begingroup$ Thank you. Now kindly have a look to the question - I have appended one more query. $\endgroup$ – hola Jul 23 '14 at 3:10

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