Geom/Trig Question M20 (p.17)

[BenCurtis]

A circle is inscribed in an equilateral triangle of side a. Show that the radius of the circle is a/√(3). [I wish to know where to begin. I drew the picture and found the tangent to the point where the radius intersected the side of the triangle to form a right triangle between that point, the top of the equilateral triangle, and the origin of the circle. I'm wondering if one of the sides of the new triangle is a/2.]

 

[Deleted member 4993]

A circle is inscribed in an equilateral triangle of side a. Show that the radius of the circle is a/√(3). [I wish to know where to begin. I drew the picture and found the tangent to the point where the radius intersected the side of the triangle to form a right triangle between that point, the top of the equilateral triangle, and the origin of the circle. I'm wondering if one of the sides of the new triangle is a/2.]

If you some name to those points:

like the vertices of the triangle is ABC, the center of the circle is O, and the tangent point where AB touches the circle is D.....

and rephrase your question wrt those points,

then I can visualize it better.

 

[soroban]

Hello, BenCurtis!

\(\displaystyle \text{A circle is inscribed in an equilateral triangle of side }a.\)
\(\displaystyle \text{Show that the radius of the circle is }\frac{a}{\sqrt{3}}\) . This is not true!

                          A
                          *
                         / \
                        /   \
                       /     \
                      /       \
                     /         \
                    /   * * *   \
                   /*           *\
               a  *               *  a
                 *                 *
                /                   \
               /*         O         *\
              / *         *         * \
             /  *         |         *  \
            /             | r           \
           /     *        |        *     \
          /       *       |       *       \
         /          *     |     *          \
      B * - - - - - - - * * * - - - - - - - * C
                a/2       D       a/2

We have equilateral triangle \(\displaystyle ABC\!:\:AB = BC = CA = a.\)
We have inscribed circle: center \(\displaystyle O\), radius \(\displaystyle r = OD,\:BD = \frac{a}{2}.\)

Draw \(\displaystyle OB.\)
Triangle \(\displaystyle ODB\) is a \(\displaystyle 30^o\!-\!60^o\) right triangle.

Therefore: .\(\displaystyle r \:=\:\dfrac{a}{2\sqrt{3}}\)

However, it is true that \(\displaystyle OA \,=\, OB \,=\, OC \,=\, \dfrac{a}{\sqrt{3}}\)

 

[BenCurtis]

Thank you Soroban. From the 30-60-90 triangle, I see how BD = √(3) = a/2. Thus a = 2√(3). And if we know that OD = r = 1, then we also know that r = a/2√(3), and by combining the equations mentioned earlier. Knowing the hypotenuse of OBC = 2 and using the information given, we find OB (and OA and OC) = a/√(3).