@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.
[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.
[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 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,
The sides of are 2x, 2y, and 2z, so its semiperimeter is x+y+z and Heron’s formula gives its area as
.
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.
The question asked for the area of the triangle, so I just substituted my values back into the area formulas.
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!