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There are given: $[BC],[DE]$, the angles $\phi_1$, $\phi_2$ and a point $A$. Build a isosceles triangle $MAN$ (with$\angle{MAN}$ = $\pi/2$) with the apex of the right angle in $A$ such as $\angle{BNC}$ =$\angle\phi_1$ and $\angle{EMD}$ =$\angle\phi_2$

thanks for your time and help!

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  • $\begingroup$ Not clear. What is $[BC]$? Is this the length of a line segment $BC$? Also, what is 'the rectangle angle'? You mean 'right angle'? Also, the points $B,C,D,E$ are fixed? or we are just given two lengths and two angles? $\endgroup$
    – mathlove
    Jan 8, 2014 at 14:50
  • $\begingroup$ Sorry for my bad math-english. $[BC]$ is the lenght of a line segment BC ( the segment itself ). There are given 2 lengths and two angles, and a point A. $\endgroup$
    – Akhtubir
    Jan 8, 2014 at 16:33
  • $\begingroup$ OK. So, the points $B,C,D,E$ are not fixed, right? $\endgroup$
    – mathlove
    Jan 8, 2014 at 16:34
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    $\begingroup$ Yes, the points are not fixed. $\endgroup$
    – Akhtubir
    Jan 8, 2014 at 16:35
  • $\begingroup$ Again not clear. What does "$\angle BNC=\phi_1, \angle EMD=\phi_2$" mean? If so, $M,N$ are fixed? I think of course not, but I don't understand it. How can we have "$\angle BNC=\phi_1, \angle EMD=\phi_2$" even if $B,C,D,E$ are not fixed? $\endgroup$
    – mathlove
    Jan 8, 2014 at 16:44

1 Answer 1

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Problem

Given the points $A$, $B$, $C$, $D$ and $E$, the line segments $BC$ and $DE$, and the angles whose measures are $\phi_1$ and $\phi_2$ ( see fig.1), find out the points $N$ and $M$ such that $m(\angle BNC) = \phi_1$, $m(\angle EMD)=\phi_2$, $m(\angle NAM) = \frac{\pi}{2}$ and $AN = AM$.

enter image description here

Solution

At first let’s recall the definition of a pair of arcos capazes:

A Pair of arcos capazes (Portuguese) is the geometrical locus of the points in the plane from which a given segment is seen under a certain angle.

For example in fig.2 is drawn a pair of arcos capazes of the line segment $DE$ under the angle whose measure is $\phi_2$.

enter image description here

The construction of a pair of arcos capazes is not difficult. See an example here: http://fr.wikipedia.org/wiki/Arc_capable

So the solution of the original problem is (See fig.4):

  1. Draw a pair of arcos capazes $\Gamma_1$ of line segment BC under an angle whose measure is $\phi_1$.

  2. Draw a pair of arcos capazes $\Gamma_2$ of line segment $DE$ under an angle whose measure is $\phi_2$.

  3. Rotate the locus $\Gamma_2$ $\frac{\pi}{2}$ rad counter-clockwise about point $A$. You will get a new locus $\Gamma_2'$.

  4. Find out the point $N$, such that $N={\Gamma_1 \cap \Gamma_2’}$.

  5. Find out the point $M$, $M \in \Gamma_2$(rotate point N $\frac{\pi}{2}$ rad clockwise about $A$).

  6. Draw the right angled isosceles $\triangle MAN$. (In our case we have two solutions $\triangle M_1AN_1$ and $\triangle M_2AN_2$).

enter image description here

Explanation of Steps 3 and 4. (See fig.3).

Let the isosceles triangle $MAN$ right-angled at $A$ , such that $M$ moves on $\Gamma_2$ and $A$ is fixed.

As the point $M$ moves on $\Gamma_2$ point N describes $\Gamma_2'$ which is nothing more than a copy of $\Gamma_2$ rotated $90$ degrees counter-clockwise relative to point $A$. So the point $N$ must be the intersection point between $\Gamma_1$ and $\Gamma_2'$.

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