Inscribed Triangle Challenge

@MathCeyhun posed an interesting geometry problem yesterday.

Even more interesting is that, as @MathCeyhun noted in a subsequent tweet, none of the posted solutions is correct.  There have been a few posted answers, but no solutions, so I thought I’d give it a try.

OBSERVATIONS

  • The perpendicular bisector of each chord was given, and perpendicular bisectors of chords always lie on radii of the circle.
  • If r is the radius of the circle, then the lengths of the extensions of the perpendicular bisectors are r-1, r-2, and r-3.
  • Nothing given guarantees anything special about the triangle, so I assumed it was scalene.  I called the side lengths 2x, 2y, and 2z to simplify the bisection labels.
  • Adding the bisector extensions, the radii to the vertices, and segment names and labels gave me this.

Inscribed_Triangle2

[Irrelevant to this problem, but I just realized by looking at this image that every triangle can be dissected into three isosceles triangles with congruent sides and a shared vertex point at the triangle’s circumcenter.  Pretty.]

SETTING UP MY SOLUTION 

Each isosceles triangle is bisected by the perpendicular bisector of its base from which I extracted three relationships from the Pythagorean theorem.

x^2+(r-1)^2=r^2 \longrightarrow x^2=(2r-1) \cdot 1

y^2+(r-2)^2=r^2 \longrightarrow y^2=(2r-2) \cdot 2

z^2+(r-3)^2=r^2 \longrightarrow z^2=(2r-3) \cdot 3

[The relationship between the half-sides, the extension of the radius beyond the triangle, and the radius is another pretty pattern.]

That gives 3 equations in 4 variables.  I needed one more to solve….

 

The area of \Delta ABC can be expressed two ways:  as the sum of the areas of the isosceles triangles, and using Heron’s formula.  From the areas of the isosceles triangles,

Area( \Delta ABC) = \frac{1}{2}(2x)(r-1) + \frac{1}{2}(2y)(r-2) + \frac{1}{2}(2z)(r-3)
Area( \Delta ABC) = x \cdot (r-1) + y \cdot (r-2) + z \cdot (r-3)

The sides of \Delta ABC are 2x, 2y, and 2z, so its semiperimeter is x+y+z and Heron’s formula gives its area as

Area( \Delta ABC) = \sqrt{(x+y+z)(-x+y+z)(x-y+z)(x+y-z)}.

The area of a given triangle is unique, so the two different area expressions are equivalent, giving a fourth equation.

SOLVING A SYSTEM & ANSWERING THE QUESTION

With four equations in four variables, I had a system of equations.  The algebra was messy, so I invoked my CAS to crunch it for me.

Inscribed_Triangle3

The question asked for the area of the triangle, so I just substituted my values back into the area formulas.

Inscribed_Triangle4

And 17.186… is clearly not one of the choices in the original problem.

 

A PLEA…

Recognizing the perpendicular bisectors, seeing all the right triangles, and connecting the multiple ways to describe the area of a triangle made this solution reasonably easy to find with the help of my computer algebra system (CAS), but I know the background algebra is, at best, cumbersome.  I hope there’s a more elegant solution, but I don’t see it.  Can anyone offer a suggestion?

Either way … this is definitely becoming a challenge problem for my Quantitative Reasoning class this coming week!

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