# Monthly Archives: September 2018

## 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.

[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.

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!

## Implicit Derivative Question

Yesterday, this question posted within the AP Calculus Community about implicit derivatives.

Below, I argue why the derivatives MUST be the same, show how four different variations can all be shown to give the same derivative, and provide a final conclusion.

INITIAL INTUITION

The Desmos graph of the given relation, is $y^2 = \frac{x-1}{x+1}$, is shown below.  Logically, it seems that even when the terms of the relation are algebraically rearranged, the graph should be invariant.   The other two forms mentioned in the Community post are on lines 2 and 3.  Lines 4, 5, and 6 show three other variations.  Here is the link to my Desmos graph allowing you to change between the forms to visually confirm the graphical invariance intuition.

If calculus “works”, it also shouldn’t matter how one calculates a derivative.   While the forms of the derivative certainly could LOOK different,  because any point on the invariant graph has the same tangent line no matter what the form of its equation, and the derivative of a relation at a point is the slope of that invariant tangent line, then the derivative also MUST be invariant.

CALCULATING “DIFFERENT” DERIVATIVES

To show the derivatives are fundamentally all the same (as suspected by the initial post), I calculate the derivatives of the equations on lines 1 and 3 given in the initial post as well as my variations on lines 4 and 6.

LINE 1:

Using the Chain Rule on the left and the Quotient Rule on the right gives

LINE 3:

This version is more complicated, requiring the Product Rule in addition to the earlier Chain and Quotient Rules.  In the penultimate line, I used the original equation to substitute for $y^2$ to transform the derivative into the same form as line 1.

LINE 4:

This time, differentiation requires only the Chain and Product Rules.

After the usual substitution for $y^2$, I multiplied both sides by $(x+1)$ to clear the denominator and solved for $y'$, returning the same result.

LINE 6:

This time, the relation is solved for x, resulting in a much more complicated Quotient+Chain Rule calculation, but substituting for $y^2$ and changing the form leads once again to the same answer.

Hopefully this is convincing evidence that all derivative forms can be shown to be equivalent.   If you’re still learning implicit differentiation, I encourage you to show the derivatives from the lines 2 and 5 variations are also equivalent.

CONCLUSION

So which approach is “best”?  In my opinion, it all depends on your personal comfort with algebraic manipulations.  Some prefer to just take a derivative from the given form of $y^2 = \frac{x-1}{x+1}$.  I avoid the more complicated quotient rule whenever I can, so the variation from line 4 would have been my approach.

The cool part is that it doesn’t matter what approach you use, so long as your algebraic manipulations are sound.  You don’t have to accept the form in which a problem is given; CHANGE IT to a form that works for you!

## Pythagorean Investigation

Here’s a challenge @jamestanton tweeted yesterday:

Visually, Tanton is asking if there is an integer right triangle (like the standard version shown on the left below) for which the integer triangle on the right exists.

The algebraic equivalent to this question is, for some $a^2+b^2=c^2$, does there exist a Natural number d so that $b^2+c^2=d^2$?

I invoked Euclid’s formula in my investigation to show that there is no value of d to make this possible.  I’d love to hear of any other proof variations.

INVOKING EUCLID’S FORMULA

For any coprime natural numbers m & n where $m>n$ and $m-n$ is odd, then every primitive Pythagorean triple can be generated by $\left\{ m^2 - n^2, 2mn, m^2 + n^2 \right\}$.

For any Natural number kevery Pythagorean triple can be generated by $\left\{ k \cdot \left( m^2 - n^2 \right), k \cdot \left( 2mn \right), k \cdot \left( m^2 + n^2 \right) \right\}$.

The generator term $k \cdot \left( m^2 + n^2 \right)$ must be the original hypotenuse (side c), but either $k \cdot \left( m^2 - n^2 \right)$ or $k \cdot \left( 2mn \right)$ can be side b.  So, if Tanton’s scenario is true, I needed to check two possible cases.  Does there exist a Natural number d such that

$d^2 = \left( k \cdot \left( m^2 - n^2 \right) \right)^2 + \left( k \cdot \left( m^2 + n^2 \right) \right)^2 = 2k^2 \left( m^4 + n^4 \right)$

or

$d^2 = \left( k \cdot \left( 2mn \right) \right)^2 + \left( k \cdot \left( m^2 + n^2 \right) \right)^2 = k^2 \left( m^4 +6m^2n^2 + n^4 \right)$

is true?

EVALUATING THE POSSIBILITIES

For the first equation, there is a single factor of 2 on the right, and there is no way to extract an odd number of factors of 2 from $\left( m^4 + n^4 \right)$ or $k^2$, so $2k^2 \left( m^4 + n^4 \right)$ can’t represent a perfect square.

For the second equation, there is no way to factor $\left( m^4 +6m^2n^2 + n^4 \right)$ over Integers, so $k^2 \left( m^4 +6m^2n^2 + n^4 \right)$ can’t be a perfect square either.

Since neither equation can create a perfect square, there is no Natural value of d that makes {b, c, d} a Pythagorean triple.  Tanton’s challenge is impossible.

Does anyone have a different approach?