# Tag Archives: proof

## Inscribed Right Angle Proof Without Words

Earlier this past week, I assigned the following problem to my 8th grade Geometry class for homework.  They had not explored the relationships between circles and inscribed angles, so I added dashed auxiliary segment AD as a hint.

What follows first is the algebraic solution I expected most to find and then an elegant transformational explanation one of my students produced.

PROOF 1:

Given circle A with diameter BC and point D on the circle.  Prove triangle BCD is a right triangle. After some initial explorations on GeoGebra sliding point D around to discover that its angle measure was always $90^{\circ}$ independent of the location of D, most successful solutions recognized congruent radii AB, AC, and AD, creating isosceles triangles CAD and BAD.  That gave congruent base angles x in triangle CAD, and y in BAD. The interior angle sum of a triangle gave $(x)+(x+y)+(y)=180^{\circ}$, or $m \angle CDB = x+y = 90^{\circ}$, confirming that BCD was a right triangle.

PROOF 2:

Then, one student surprised us.  She marked the isosceles base angles as above before rotating $\Delta BCD$ $180^{\circ}$ about point A. Because the diameter rotated onto itself, the image and pre-image combined to form an quadrilateral with all angles congruent.  Because every equiangular quadrilateral is a rectangle, M had confirmed BCD was a right triangle.

CONCLUSION:

I don’t recall seeing M’s proof before, but I found it a delightfully elegant application of quadrilateral properties.  In my opinion, her rotation is a beautiful proof without words solution.

Encourage freedom, flexibility of thought, and creativity, and be prepared to be surprised by your students’ discoveries!

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## Quadratics + Tangent = ???

Here’s a very pretty problem I encountered on Twitter from Mike Lawler 1.5 months ago.

I’m late to the game replying to Mike’s post, but this problem is the most lovely combination of features of quadratic and trigonometric functions I’ve ever encountered in a single question, so I couldn’t resist.  This one is well worth the time for you to explore on your own before reading further.

My full thoughts and explorations follow.  I have landed on some nice insights and what I believe is an elegant solution (in Insight #5 below).  Leading up to that, I share the chronology of my investigations and thought processes.  As always, all feedback is welcome.

WARNING:  HINTS AND SOLUTIONS FOLLOW

Investigation  #1:

My first thoughts were influenced by spoilers posted as quick replies to Mike’s post.  The coefficients of the underlying quadratic, $A^2-9A+1=0$, say that the solutions to the quadratic sum to 9 and multiply to 1.  The product of 1 turned out to be critical, but I didn’t see just how central it was until I had explored further.  I didn’t immediately recognize the 9 as a red herring.

Basic trig experience (and a response spoiler) suggested the angle values for the tangent embedded in the quadratic weren’t common angles, so I jumped to Desmos first.  I knew the graph of the overall given equation would be ugly, so I initially solved the equation by graphing the quadratic, computing arctangents, and adding. Insight #1:  A Curious Sum

The sum of the arctangent solutions was about 1.57…, a decimal form suspiciously suggesting a sum of $\pi/2$.  I wasn’t yet worried about all solutions in the required $[0,2\pi ]$ interval, but for whatever strange angles were determined by this equation, their sum was strangely pretty and succinct.  If this worked for a seemingly random sum of 9 for the tangent solutions, perhaps it would work for others.

Unfortunately, Desmos is not a CAS, so I turned to GeoGebra for more power.

Investigation #2:

In GeoGebra, I created a sketch to vary the linear coefficient of the quadratic and to dynamically calculate angle sums.  My procedure is noted at the end of this post.  You can play with my GeoGebra sketch here.

The x-coordinate of point G is the sum of the angles of the first two solutions of the tangent solutions.

Likewise, the x-coordinate of point H is the sum of the angles of all four angles of the tangent solutions required by the problem. Insight #2:  The Angles are Irrelevant

By dragging the slider for the linear coefficient, the parabola’s intercepts changed, but as predicted in Insights #1, the angle sums (x-coordinates of points G & H) remained invariant under all Real values of points A & B.  The angle sum of points C & D seemed to be $\pi/2$ (point G), confirming Insight #1, while the angle sum of all four solutions in $[0,2\pi]$ remained $3\pi$ (point H), answering Mike’s question.

The invariance of the angle sums even while varying the underlying individual angles seemed compelling evidence that that this problem was richer than the posed version.

Insight #3:  But the Angles are bounded

The parabola didn’t always have Real solutions.  In fact, Real x-intercepts (and thereby Real angle solutions) happened iff the discriminant was non-negative: $B^2-4AC=b^2-4*1*1 \ge 0$.  In other words, the sum of the first two positive angles solutions for $y=(tan(x))^2-b*tan(x)+1=0$ is $\pi/2$ iff $\left| b \right| \ge 2$, and the sum of the first four solutions is $3\pi$ under the same condition.  These results extend to the equalities at the endpoints iff the double solutions there are counted twice in the sums.  I am not convinced these facts extend to the complex angles resulting when $-2.

I knew the answer to the now extended problem, but I didn’t know why.  Even so, these solutions and the problem’s request for a SUM of angles provided the insights needed to understand WHY this worked; it was time to fully consider the product of the angles.

Insight #4:  Finally a proof

It was now clear that for $\left| b \right| \ge 2$ there were two Quadrant I angles whose tangents were equal to the x-intercepts of the quadratic.  If $x_1$ and $x_2$ are the quadratic zeros, then I needed to find the sum A+B where $tan(A)=x_1$ and $tan(B)=x_2$.

From the coefficients of the given quadratic, I knew $x_1+x_2=tan(A)+tan(B)=9$ and $x_1*x_2=tan(A)*tan(B)=1$.

Employing the tangent sum identity gave $\displaystyle tan(A+B) = \frac{tan(A)+tan(B)}{1-tan(A)tan(B)} = \frac{9}{1-1}$

and this fraction is undefined, independent of the value of $x_1+x_2=tan(A)+tan(B)$ as suggested by Insight #2.  Because tan(A+B) is first undefined at $\pi/2$, the first solutions are $\displaystyle A+B=\frac{\pi}{2}$.

Insight #5:  Cofunctions reveal essence

The tangent identity was a cute touch, but I wanted something deeper, not just an interpretation of an algebraic result.  (I know this is uncharacteristic for my typically algebraic tendencies.)  The final key was in the implications of $tan(A)*tan(B)=1$.

This product meant the tangent solutions were reciprocals, and the reciprocal of tangent is cotangent, giving $\displaystyle tan(A) = \frac{1}{tan(B)} = cot(B)$.

But cotangent is also the co-function–or complement function–of tangent which gave me $tan(A) = cot(B) = tan \left( \frac{\pi}{2} - B \right)$.

Because tangent is monotonic over every cycle, the equivalence of the tangents implied the equivalence of their angles, so $A = \frac{\pi}{2} - B$, or $A+B = \frac{\pi}{2}$.  Using the Insights above, this means the sum of the solutions to the generalization of Mike’s given equation, $(tan(x))^2+b*tan(x)+1=0$ for x in $[0,2\pi ]$ and any $\left| b \right| \ge 2$,

is always $3\pi$ with the fundamental reason for this in the definition of trigonometric functions and their co-functions.  QED

Insight #6:  Generalizing the Domain

The posed problem can be generalized further by recognizing the period of tangent: $\pi$.  That means the distance between successive corresponding solutions to the internal tangents of this problem is always $\pi$ each, as shown in the GeoGebra construction above.

Insights 4 & 5 proved the sum of the angles at points C & D was $\pi/2$.  Employing the periodicity of tangent,  the x-coordinate of $E = C+\pi$ and $F = D+\pi$, so the sum of the angles at points E & F is $\frac{\pi}{2} + 2 \pi$.

Extending the problem domain to $[0,3\pi ]$ would add $\frac{\pi}{2} + 4\pi$ more to the solution, and a domain of $[0,4\pi ]$ would add an additional $\frac{\pi}{2} + 6\pi$.  Pushing the domain to $[0,k\pi ]$ would give total sum $\displaystyle \left( \frac{\pi}{2} \right) + \left( \frac{\pi}{2} +2\pi \right) + \left( \frac{\pi}{2} +4\pi \right) + \left( \frac{\pi}{2} +6\pi \right) + ... + \left( \frac{\pi}{2} +2(k-1)\pi \right)$

Combining terms gives a general formula for the sum of solutions for a problem domain of $[0,k\pi ]$ $\displaystyle k * \frac{\pi}{2} + \left( 2+4+6+...+2(k-1) \right) * \pi =$ $\displaystyle = k * \frac{\pi}{2} + (k)(k-1) \pi =$ $\displaystyle = \frac{\pi}{2} * k * (2k-1)$

For the first solutions in Quadrant I, $[0,\pi]$ means k=1, and the sum is $\displaystyle \frac{\pi}{2}*1*(2*1-1) = \frac{\pi}{2}$.

For the solutions in the problem Mike originally posed, $[0,2\pi]$ means k=2, and the sum is $\displaystyle \frac{\pi}{2}*2*(2*2-1) = 3\pi$.

I think that’s enough for one problem.

APPENDIX

My GeoGebra procedure for Investigation #2:

• Graph the quadratic with a slider for the linear coefficient, $y=x^2-b*x+1$.
• Label the x-intercepts A & B.
• The x-values of A & B are the outputs for tangent, so I reflected these over y=x to the y-axis to construct A’ and B’.
• Graph y=tan(x) and construct perpendiculars at A’ and B’ to determine the points of intersection with tangent–Points C, D, E, and F in the image below
• The x-intercepts of C, D, E, and F are the angles required by the problem.
• Since these can be points or vectors in Geogebra, I created point G by G=C+D.  The x-intercept of G is the angle sum of C & D.
• Likewise, the x-intercept of point H=C+D+E+F is the required angle sum.

## Unanticipated Proof Before Algebra

I was talking with one of our 5th graders, S,  last week about the difference between showing a few examples of numerical computations and developing a way to know something was true no matter what numbers were chosen.  I hadn’t started our conversation thinking about introducing proof.  Once we turned in that direction, I anticipated scaffolding him in a completely different direction, but S went his own way and reinforced for me the importance of listening and giving students the encouragement and room to build their own reasoning.

SETUP:  S had been telling me that he “knew” the product of an even number with any other number would always be even, while the product of any two odds was always odd.  He demonstrated this by showing lots of particular products, but I asked him if he was sure that it was still true if I were to pick some numbers he hadn’t used yet.  He was.

Then I asked him how many numbers were possible to use.  He promptly replied “infinite” at which point he finally started to see the difficulty with demonstrating that every product worked.  “We don’t have enough time” to do all that, he said.  Finally, I had maneuvered him to perhaps his first ever realization for the need for proof.

ANTICIPATION:  But S knew nothing of formal algebra.  From my experiences with younger students sans algebra, I thought I would eventually need to help him translate his numerical problem into a geometric one.  But this story is about S’s reasoning, not mine.

INSIGHT:  I asked S how he would handle any numbers I asked him to multiply to prove his claims, even if I gave him some ridiculously large ones.  “It’s really not as hard as that,” S told me.  He quickly scribbled on his paper and covered up all but the one’s digit.  “You see,” he said, “all that matters is the units.  You can make the number as big as you want and I just need to look at the last digit.”  Without using this language, S was venturing into an even-odd proof via modular arithmetic.

With some more thought, he reasoned that he would focus on just the units digit through repeated multiples and see what happened.

FIFTH GRADE PROOF:  S’s math class is currently working through a multiplication unit in our 5th grade Bridges curriculum, so he was already in the mindset of multiples.  Since he said only the units digit mattered, he decided he could start with any even number and look at all of its multiples.  That is, he could keep adding the number to itself and see what happened.  As shown below, he first chose 32 and found the next four multiples, 64, 96, 128, and 160.  After that, S said the very next number in the list would end in a 2 and the loop would start all over again. He stopped talking for several seconds, and then he smiled.  “I don’t have to look at every multiple of 32.  Any multiple will end up somewhere in my cycle and I’ve already shown that every number in this cycle is even.  Every multiple of 32 must be even!”  It was a pretty powerful moment.  Since he only needed to see the last digit, and any number ending in 2 would just add 2s to the units, this cycle now represented every number ending in 2 in the universe.  The last line above was S’s use of 1002 to show that the same cycling happened for another “2 number.”

DIFFERENT KINDS OF CYCLES:  So could he use this for all multiples of even numbers?  His next try was an “8 number.” After five multiples of 18, he achieved the same cycling.  Even cooler, he noticed that the cycle for “8 numbers” was the 2 number” cycle backwards.

Also note that after S completed his 2s and 8s lists, he used only single digit seed numbers as the bigger starting numbers only complicated his examples.  He was on a roll now. I asked him how the “4 number” cycle was related.  He noticed that the 4s used every other number in the “2 number” cycle.  It was like skip counting, he said.  Another lightbulb went off.

“And that’s because 4 is twice 2, so I just take every 2nd multiple in the first cycle!”  He quickly scratched out a “6 number” example. This, too, cycled, but more importantly, because 6 is thrice 2, he said that was why this list used every 3rd number in the “2 number” cycle.  In that way, every even number multiple list was the same as the “2 number” list, you just skip-counted by different steps on your way through the list.

When I asked how he could get all the numbers in such a short list when he was counting by 3s, S said it wasn’t a problem at all.  Since it cycled, whenever you got to the end of a list, just go back to the beginning and keep counting.  We didn’t touch it last week, but he had opened the door to modular arithmetic.

I won’t show them here, but his “0 number” list always ended in 0s.  “This one isn’t very interesting,” he said.  I smiled.

ODDS:  It took a little more thought to start his odd number proof, because every other multiple was even.  After he recognized these as even numbers, S decided to list every other multiple as shown with his “1 number” and “3 number” lists. As with the evens, the odd number lists could all be seen as skip-counted versions of each other.  Also, the 1s and 9s were written backwards from each other, and so were the 3s and 7s.  “5 number” lists were declared to be as boring as “0 numbers”.  Not only did the odds ultimately end up cycling essentially the same as the evens, but they had the same sort of underlying relationships.

CONCLUSION:  At this point, S declared that since he had shown every possible case for evens and odds, then he had shown that any multiple of an even number was always even, and any odd multiple of an odd number was odd.  And he knew this because no matter how far down the list he went, eventually any multiple had to end up someplace in his cycles.  At that point I reminded S of his earlier claim that there was an infinite number of even and odd numbers.  When he realized that he had just shown a case-by-case reason for more numbers than he could ever demonstrate by hand, he sat back in his chair, exclaiming, “Whoa!  That’s cool!”

It’s not a formal mathematical proof, and when S learns some algebra, he’ll be able to accomplish his cases far more efficiently, but this was an unexpectedly nice and perfectly legitimate numerical proof of even and odd multiples for an elementary student.

## Mixed Number Curiosity

The first part of this is not my work, but I offer an intriguing extension.

PROBLEM:

This appeared on Twitter recently. (source) Despite its apparent notional confusion, it is a true statement.  Since both sides are positive, you can square both sides without producing extraneous results.  Doing so proves the statement.

It’s a lovely, but curious piece of arithmetic trivia.  A more mathematical question:

Does this pattern hold for any other numbers?

Thomas Oléron Evans has a proof on his ‘blog here in which he solves the equation $\sqrt(a+\frac{b}{c}) = a \cdot \sqrt( \frac{b}{c})$ under the assumptions that a, b, and c are natural and $\frac{b}{c}$ is any fraction in its most reduced form.  Doing so leads to the equation where A is any natural number larger than 1.  Nice.

While the derivation is more complicated for middle and upper school students, proof that the formula works is straightforward.

A>0, so all terms are positive.  Square both terms, find a common denominator, et voila!

EXTENSIONS:

Using Evans’ assumptions, the formula is inevitable, but any math rests on its assumptions.  I wondered if there are more numbers out there for which the original number pattern was true.

Using Evans’ formula, my very first thought was to violate the integer assumption.  I let $A=1.1$ and grabbed my Nspire. Checking the fractional term, I see that I also violated the “simplest form” assumption.  Converting this to a fractional form to make sure there isn’t a decimal off somewhere down the line, I got So it is true for more than Evans claimed.

I don’t have time to investigate this further right now, so I throw it out to you.  How far does this property go?

## Tell a Friend

I’ve been in several conversations over these first couple weeks of school with colleagues in our lower and middle schools about what students need to do to convince others they understand an idea.

On our first pre-assessments, some teachers noted that many students showed good computation skills, but struggled when they had to explain relationships.  Frankly, I’m never surprised by revelations that students find explanations more difficult than formulas and computations.  That’s tough for learners of all ages.  But, in my opinion, it’s also the most important part about developing a way to communicate mathematically.

In the other direction, I frequently hear students complain that they just don’t know what to write and that teachers seem to arbitrarily ask for “more explanation”, but they just can’t figure out what that means.

SOLUTION?:

Just like writing in humanities classes, a math learner needs to seriously consider his “audience”.  Who’s going to read your solution?  I think too many write for a classroom teacher, expecting him or her to fill in any potential logical gaps.

Instead, I tell my students that I expect all of their explanations to be understandable by every classmate. In short,

Don’t write your answer to me; write it to a friend who’s been absent for a couple days.

If a random classmate who’s been out a couple days can get it just based on your written work, they you’re good.

## Lovely or Tricky Triangle Question?

In addition to not being drawn to scale and asking for congruence anyway, I like this problem because it potentially forces some great class discussions.

One responder suggested using the Law of Sines (LoS) to establish an isosceles triangle.  My first thought was that was way more sophisticated than necessary and completely missed the fact that the given triangle information was SSA.

My initial gut reaction was this SSA setup was a “trick” ambiguous case scenario and no congruence was possible, but I couldn’t find a flaw in the LoS logic. After all, LoS fails when attempting to find obtuse angles, but the geometry at play here clearly makes angles B and C both acute.  That meant LoS should work, and this was actually a determinate SSA case, not ambiguous.  I was stuck in a potential contradiction.  I was also thinking with trigonometry–a far more potent tool than I suspected was necessary for this problem.

“Stuck” moments like this are GOLDEN for me in the classroom.  I could imagine two primary student situations here.  They either  1) got a quick “proof” without recognizing the potential ambiguity, or 2) didn’t have a clue how to proceed.  There are many reasons why a student might get stuck here, all of which are worth naming and addressing in a public forum.  How can we NAME and MOVE PAST situations that confuse us?  Perhaps more importantly, how often do we actually recognize when we’re in the middle of something that is potentially slipperier than it appears to be on the surface?

PROBLEM RESOLUTION:

I read later that some invoked the angle bisector theorem, but I took a different path.  I’m fond of a property I asked my geometry classes to prove last year .

If any two of a triangle’s 1) angle bisector, 2) altitude, and 3) median coincide, prove that the remaining segment does, too, and that whenever this happens, the triangle will be isosceles with its vertex at the bisected angle.

Once I recognized that the angle bisector of angle BAC was also the median to side BC, I knew the triangle was isosceles.  The problem was solved without invoking any trigonometry or any similarity ratios.

Very nice problem with VERY RICH discussion potential.  Thanks for the tweet, Mr. Noble.

For more conversation on this, check out this Facebook conversation.

## Unexpected Proof of the Pythagorean Theorem

Following is a neat discovery of an alternative proof of the Pythagorean Theorem resulting from the multiple solutions to the Circle and Square problem.  I’m sure someone has seen this before, as there are literally 100s of unique proofs of the Pythagorean Theorem, but this one was new to me.

The intersecting chord relationships in a circle can be proven using only similar triangles.  Proofs of these are at the bottom of this post, if needed.  Using only those, you can prove the Pythagorean Theorem.

PROOF:

The image below–a revision of the diagram from my previous post–shows diameter DE in circle C.  Chord AB is a side of the given square from the Circle and Square problem and is bisected by symmetry into two segments, each of length a.  Let  be the radius of circle C.  Let the portion of DE from point C to chord AB have length b.  Because AB is a chord bisected by diameter DE, two right triangles are created, as shown. AB and DE are intersecting chords, so $a \cdot a = (r-b) \cdot (r+b)$.  Expanding the right side and moving the $b^2$ term to the other side gives the Pythagorean Theorem.

Short and sweet once the chord relationships are established.

SUPPORTING PROOF 1:

In the image below, AB and CD are any two chords intersecting at point E.  Vertical angles give $\angle DEA \cong \angle BEC$.  Because $\angle ADE$ and $\angle CBE$ are inscribed angles sharing arc AC, they are also congruent. That means $\Delta ADE \sim \Delta CBE$, which gives $\displaystyle \frac{x}{w} = \frac{y}{z}$, or $x \cdot z = w \cdot y$.  QED

SUPPORTING PROOF 2:
Show that if a diameter bisects a chord, the diameter and chord are perpendicular.  Start with the vertical diameter of circle C bisecting chord AB. It should be straightforward to show $\Delta ADC \cong \Delta BDC$ by SSS.  That means  corresponding angles $\angle ADC \cong \angle BDC$; as they also from a linear pair, those angles are both right, and the proof is established.