Math: My Solutions to mathhombre’s Problems

Part of the job of a teacher is to model her or his Subject-specific thinking to help students understand the subject techniques and the Subject processes, patterns and metacontent. To this end, I dedicate this post to modelling how I tackled the math problems, created by John Golden as part of a math final for preservice teachers, that I mentioned in my last post.

mathhombre's Twitter avatar I enjoy these types of problems. They are chuck full of math — and Math — from geometry and trigonometry to algebra and arithmetic. But even more so, they require mathematical thought, imagination, creativity, logic and perseverance to solve them. Had John expanded the questions to ask for metacognitive reflection and recording, at least the trigonometry question could easily pass as a Math lab.

Once again I invite you to visit his post and take a crack at his problems before reading on.

The remainder of this post contains my solutions to these problems. (I went overboard with the trigonometric problem. I blame it on play and slightly improved health.)

Trig Problem 2

The first problem John offers is a circle geometry / trigonometry problem. He presents a diagram and simply asks the test taker to “Figure out some of the missing information in the diagram”. The only information the student is sure of from this task is that some information — more than one piece — is missing. Interpretation, logic and problem solving are all the tools the student has to continue with the question.

As I mentioned in my last post, what I really like about this question — and the next one — is its openness. The student isn’t asked for all possible missing information. He can choose what she wants to explore and discover.

I began my activity to this task by stating two assumptions on which the rest of my work rests. I assumed that point C is the center of the big circle. I also assumed that point A is the center of the small one. You might think this obvious given the diagram, but I like stating these assumptions right away. That way everyone knows where I stand and the rest of my work can not be faulted based on ambiguity on these points.

I then stated the explicit givens, that angle ADE is 30°, angles AED and ACD are 90°, and segment AC is 3 units long. (All of this and the rest of the original problem diagram is in light blue in the answer diagram below.)

Here is a diagram of my answer.



Click to enlarge or visit the GeoGebra construct from which it derives.

Joy, Pythagoras

Based on the assumptions that C is the center of the big circle and A is the center of the small one, I calculated and labelled the radii for the big and small circles. So CD, CI, CL and CM are each 6 units long and AC, AG, AH, AI, AJ and AK are each 3 units long.

This made apparent that triangle ADC is a right triangle with legs CD of 6 units and AC of 3 units, allowing (through Pythagoras’s Theorem) hypotenuse AD to be calculated as square root 45 or 3√5, which is also the hypotenuse of triangle AED.

Making angles

Equally obvious to the use of Pythagoras’s Theorem to calculate the common hypotenuse of triangles ADC and AED, is the use of the 180° (or π) “rule” for use in triangle AED.

This concept is so well known, it has almost become pneumonic. Yet, think on how profound the notion that the internal angles of all triangles always add up to a common constant, and that this constant, 180°, is also found in: the angles that accumulate to make a line, the conversion from degrees into radians and the command that sends us back the way we came.

In triangle AED, angle DAE is 60° because angle AED is 90° and angle EDA is 30° and these add up to 180°.

Attempting a similar algorithm in triangle ADC requires some labelling. I set the measure of angle ADC to unknown x, which makes the measure of angle CAD to 90-x by the 180° rule. But the fun does not stop there. Since the angles touching line EF on one side at common point A also add up to 180°, angle FAC is 30+x (60+(90-x)+(30+x)=180).

One could circle point A applying this rule to label all angles surrounding A, but another way is to apply the Vertically Opposite Angle Rule (yet another pneumonic one can ignore since the 180° rule produces it) to determine that angle JAK is 60°, IAJ is 90-x and HAI is 30+x.

Similar rules can be used to determine that all the angle measurements about points C and E are 90°, but also to calculate that angles CFA, LFN, ABE and VBW are 60-x in measure. (Oh, by the way, I added points B, L, N, M, P, Q, R, V, W, S, T and U to the diagram. The last five are hidden off the edge, but are used solely to allow labelling of angles VBW, UDT, TDS and LFN vertically opposite of and equal in measure to angles ABE, EDA, ADC and CFA respectively.)

The value of x, and consequently of the other angles, is calculated applying the Law of Cosines in triangle ADC.

AC² = AD² + CD² – 2(AD)(CD)cos(x)
3² = (3√5)² + 6² – 2(3√5)(6)cos(x)
9 = 45 + 36 – (36√5)cos(x)
cos(x) = [(-72)/(-36)](1/√5)
cos(x) = 2/√5
x = 26.57°

From this, 90-x = 63.43°, 30+x = 56.57° and 60-x = 33.43°. Using the 180° rule to back check these values confirms them.

These calculations take care of all the obvious angles that the problem suggests are missing.

Measuring sides

Observation of the diagram to this point reveals that quadrilateral AEDC is not a kite since AC ≠ AE (AH = AC and AE = AH + EH), angle DAE (at 60°) ≠ CAD (at 63.43°) and angle EDA (at 30°) ≠ ADC (at 26.57°).

The lengths of AE and DE need to be calculated.

I used the Law of Sines to calculate the lengths of AE and DE. This is fairly easy since the sines of 30°, 60° and 90° produce easy ratios, 1/2, √3/2 and 1 respectively. The calculation looks like this.

sin(30°)/AE = sin(60°)/DE = sin(90°)/(3√5)
1/(2AE) = √3/(2DE) = 1/(3√5)
AE = 3√5/2, DE = 3√15/2

I repeated this process to calculate the lengths of AF, CF, BA and BE.

sin(33.43°)/3 = sin(90°)/(AF) = sin(56.57°)/(CF)
AF = 3sin(90°)/sin(33.43°) = 5.45 units
CF = 3sin(56.57°)/sin(33.34°) = 4.54 units

sin(33.43°)/(3√5/2) = sin(90°)/(BA) = sin(56.57°)/(BE)
BA = 3√5sin(90°)/2sin(33.43°) = 6.09 units
BE = 3√5sin(56.57°)/2sin(33.43°) = 5.08 units

It did not escape me that triangles ABE and ACF, with equivalent interior angles, are mathematically similar.

Having calculated the length of CF as 4.54 units and the length of CL as 6 units, it holds that FL = CL – CF = 1.46 units.

Unfortunately, the same can not be said for calculating the lengths of EQ, EP and FN, which remain undetermined. Perhaps there is a trick to calculating them. I was considering the power of a point using secants BD (through Q) and BM (through I) to calculate EQ, but the lengths of DQ, BQ, MI and BI are unknown. Unfortunately, the length of “chord” RQ, subtended by angle EDA and creating triangle QDR, also can not be determined because the length of DQ is again unknown. With point A and angle DAE mathematically set, the length of chord PN might be calculable, paving the way to calculating the lengths of EP and FN. I have not yet figured out how to do this. This leaves EQ, EP and FN unmeasured.

Measuring JR

The length of JR is 1.02 units. But I cheated to determine this!

Whereas all the other determined values described in this post are calculable from the initial information provided in this post and John’s original problem, calculating the length of JR required mechanical measurement of another length, mainly the length of DO. GeoGebra did the measuring, and had it not done so, the length of JR would be undetermined like those of EQ, EP and FN.

Wait! So why didn’t I measure the lengths of EQ, EP and FN? For the lengths of these segments, I have no theory to back my measurements with. For calculating the length of JR, I do.

The length of JR can be calculated using the Side-Splitter Theorem. (Funny, I used to call this the Parallel Projection Theorem. The Side-Splitter Theorem sounds like a comedy act.)

Segment AC splits triangle DOR parallel to leg OR of that triangle. The proportions of triangle DCA are already known to be CD:AC:AD = 6:3:3√5. So knowing the length of DO or OR can provide the length of DR in triangle DOR.

Here is where the mechanical measurement comes in. I used GeoGebra to measure (not calculate) the length of DO at 9.60 units. From this, using proportional scaling of similar triangles DCA and DOR, I calculate the lengths of OR and DR as 4.80 units and 10.73 units respectively.

The length of JR = DR – DA – AJ = 10.73 – 3√5 – 3 = 1.02 units.

Still if this were a real exam and I did not have access to GeoGebra, I would not be able to determine JR’s length.

Why stop there?

That essentially covers the basic values of John’s original problem that any preservice teacher would reasonably be expected to determine on a test. But why stop there?

I am under no time pressure and I saw right from the start that I can calculate values associated with those circles. I am sure John did not expect such a step, but here I am noticing patterns and recognizing opportunities. So, here John are my bonus marks.

The circumference (2π*radius) of the small circle is 6π since its radius is 3 units. That of the big circle is 12π from its radius of 6 units.

The central angles of these circles produce isosceles triangles with radial legs of either 3 units (small circle) or 6 units (big circle) and chords subtended by these angles.

In particular, the small circle’s central angles GAH and JAK at 60° suggest equilateral triangles AHG and AKJ. They also suggest that arcs GH and JK have arclengths of π. Since these triangles are equilateral, GH and JK are 3 units long just like the radius of the circle. This implies an arclength of one radian or π. Independently, 60° = (60°/180°)π = π/3, while the arclength of an arc of a circle is calculated as the measure of its central angle times the radius of the circle (angle*radius). In this case angle*radius = (π/3)*(3) = π.

Since the other central angles of the small circle are not 60°, their angles times three are used to calculate the arclengths of their arcs and the Law of Cosines is used to calculate the lengths of their chords.

The length of CG, for instance, is 3.15 units as calculated below.

(CG)² = (AC)² + (AG)² – 2(AC)(AG)cos(angle CAG)
(CG)² = 3² + 3² – 2(3)(3)cos(63.43°)
CG = 3.15 units

Arc CG, corresponding to this chord and similarly subtended by central angle CAG, has arclength 1.06π units ((63.43°/180°)*3 = 1.06).

For the big circle, there are four isosceles triangles with 6 unit-long radial legs and 6√2 unit-long chords (calculated using Pythagoras’s Theorem since the central angles around C are 90°). The corresponding arcs are all 3π in arclength.

I used the 180° rule for triangles, used in the Making Angles section, to calculate the other angles of each of these isosceles triangles.

For triangle CAG, angles AGC and GCA are each equal to 180° – angle CAG = 180° – 64.43° = 58.29° in measure.

And that exhausts the values I calculated.

What I missed

I noticed upon reviewing my answer diagram that I forgot to calculate the “exterior angles” around B, D and F, that is angles WBI, EBV, NFC, KFL, CDU and SDE. However, I did calculate these in the GeoGebra construct using the 180° rule for lines.

Angle WBI = angle EBV = 180° – 33.43° = 146.57°
Angle NFC = angle KFL = 180° – 33.43° = 146.57°

Angle SDE = angle CDU
180° – angle TDS – angle EDA
180° – angle ADC – angle UDT
180° – 26.57° – 30°
Angle SDE = angle CDU = 123.43°

Geometry Problem 1

Relative to the circle geometry / trigonometry problem above, John’s line geometry problem is trivial. There is not much to do with it except fill in the missing angles following the given information.

I started out by assuming that lines JK and LM were parallel to each other. This was a pretty good assumption, given that John was testing student-teacher knowledge of parallel lines, perpendicular lines, transversal lines and associated angles.

Here is my answer. I will leave the explanation to you.



Click to enlarge or visit the GeoGebra construct from which it derives.

Benefitting Our Students

Solving an unfamiliar problem in front of, or better yet, with our students can greatly benefit them. They get to see how an unfamiliar problem is solved. They get to see how we approach and think on a problem. They get clearer — or messier — explanations as we think aloud. They even get to see us make mistakes and how we recognize and resolve them.

Sometimes modelling mathematical activity when faced when an unfamiliar task or challenge can flop on us and backfire on our kids. But how we deal with these flops is also informative.

Most of the time, our flops are minor — and often silly (yes, 1 + 2 = 3, not 4) — and our modelling, after correcting the error, leads to a correct answer.

But what our students gain is our thinking, reasoning and problem solving expertise. They learn the difference between solving and solution, between proving and proof. They learn to be — and how to be — creative, logical, persistent and progressive in our tackling of a problem. They learn to touch Math, instead of just math.

And that is where we want our students to be.

Gone are the days where the answer to a particular problem is satisfactory. Now thinking is where it’s at. We have to model our thinking and discuss solving with our students. We have to teach them to think, so they can excel beyond our meager abilities.

Our students deserve better and modelling is one key way (another would be rich math problems and labs) to ensure they get the meta out of their math.


2 thoughts on “Math: My Solutions to mathhombre’s Problems

  1. Pingback: Math: Let Students Earn Their Weight « Digital Substitute

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