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Circles Tangent to an Isosceles Triangle

tncirtri.html

Exploration

A circle is inscribed in an isosceles triangle with sides a, a and 2b in length.  A second, smaller circle is inscribed tangent to the first circle and to the equal sides of the triangle.  Show that the radius of the second circle is

"tncirtri_1.gif".

Assume that a>b.

Approach

Construct an isosceles triangle whose sides are the given lengths.  Construct the circle inscribed in the triangle.  The point of tangency between the first and second circle is at the parameter θ=π/2 on the first circle.  Construct a second triangle from the equal-length sides and a line tangent to the first circle at the tangency point.  The second circle can then be inscribed inside the second triangle.  Find and simplify the radius of the second inscribed circle.

Initialize

To initialize Descarta2D, select the input cell bracket and press SHIFT-Enter.

This initialization assumes that the Descarta2D software has been copied into one of the standard directories for AddOns which are on the Mathematica search path, $Path.

<<Descarta2D`

Solution

Construct the isosceles triangle with the proper side lengths.

Clear[a,b];
T1=Triangle2D[{a,a,2b}] //FullSimplify

"tncirtri_2.gif"

Construct the first inscribed circle and simplify the result.

Clear[E1,E2];
C1=(Circle2D[T1,Inscribed2D] //FullSimplify) //.
      {Sqrt[a^2]->a, Sqrt[b^2]->b,
       Sqrt[E1_*b^2/E2_]->b*Sqrt[E1/E2]} //FullSimplify

"tncirtri_3.gif"

Construct the point of tangency between the first circle and the second.

P1=Point2D[C1[Pi/2]] //FullSimplify

"tncirtri_4.gif"

Construct the second triangle.  The results are complicated, so we define and use some simplification rules that are applied to the result.

rules1={
   -1+2a/(a+b)->(a-b)/(a+b),
   Sqrt[(a-b)(a+b)]/(a+b)->Sqrt[(a-b)]/Sqrt[(a+b)],
   Sqrt[(a-b)/(a+b)]->Sqrt[a-b]/Sqrt[a+b],
   1/Sqrt[a^2-b^2]->1/(Sqrt[a-b]*Sqrt[a+b]),
   Sqrt[a^2-b^2]->Sqrt[a-b]*Sqrt[a+b]};
L1=Line2D[Segment2D[T1,2,3]] //FullSimplify;
L2=Line2D[Segment2D[T1,1,3]] //FullSimplify;
T2=Triangle2D[L1,L2,Line2D[P1,0]];
T2=(T2 //. rules1 //Simplify) //. rules1

"tncirtri_5.gif"

Construct the circle inscribed in the second triangle and find the radius.  The results are complicated, so we define and use some simplification rules that are applied to the result.

rules2={
   Sqrt[a^2*(a-b)^2/(a+b)^2]->a(a-b)/(a+b),
   Sqrt[(a-b)^2*b^2/(a+b)^2]->b(a-b)/(a+b),
   Sqrt[(a-b)^3*b^2/(a+b)^3]->b*Sqrt[(a-b)^3/(a+b)^3]};
R2=((Radius2D[C2=Circle2D[T2,Inscribed2D]] // Simplify) //.
      rules2 // Simplify) //. rules2

"tncirtri_6.gif"

Discussion

This is the plot of a numerical example with a=10 and b=3.5.

Sketch2D[{T1,C1,P1,T2,C2} /. {a->10,b->3.5}]

"tncirtri_7.gif"

Graphics saved as "tncirt01.eps".

This is another example with a=10 and b=6.

Sketch2D[{T1,C1,P1,T2,C2} /. {a->10,b->6}]

"tncirtri_8.gif"

Graphics saved as "tncirt02.eps".


Copyright © 1999-2007 Donald L. Vossler, Descarta2D Publishing
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