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A lovely angle puzzle

March 24, 2016 4 comments

I’ve written before about the app “Brilliant“, which is well worth getting, and I also follow their Facebook page which provides me with a regular stream questions. Occasionally I have to think about how to tackle them, and they’re excellent. More often, a question comes up that I look at and think would be awesome to use in a lesson.

Earlier this week this question popped up:

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What a lovely question that combines algebra and angle reasoning! I can’t wait to teach this next time, and I am planning on using this as a starter with my y11 class after the break.

The initial question looks simple, it appears you sum the angles and set it equal to 360 degrees, this is what I expect my class to do. If you do this you get:

7x + 2y + 6z – 20 = 360

7x + 2y + 6z = 380 (1)

I anticipate some will try to give up at this point, but hopefully the resilience I’ve been trying to build will kick in and they’ll see they need more equations. If any need a hint I will tell them to consider vertically opposite angles. They should then get:

2x – 20 = 2y + 2z (2)

And

3x = 2x + 4z (3)

I’m hoping they will now see that 3 equations and 3 unknowns is enough to solve. There are obviously a number of ways to go from here. I would rearrange equation 3 to get:

x = 4z (4)

Subbing into 2 we get:

8z – 20 = 2y + 2z

6z = 2y + 20 (5)

Subbing into 1

28z + 2y + 6z = 380

34z = 380 – 2y (6)

Add equation  (5) to (6)

40z = 400

z = 10 (7)

Then equation 4 gives:

x = 40

And equation 2 gives:

60 = 2y + 20

40 = 2y

y = 20.

From here you can find the solution x + y + z = 40 + 20 + 10 = 70.

A lovely puzzle that combines a few areas and needs some resilience and perseverance to complete. I enjoyed working through it and I’m looking forward to testing it out on some students.

Cross-posted to Betterqs here.

A puzzle with possibilities

December 9, 2015 Leave a comment

Brilliant’s Facebook page is a fantastic source of brain teasers, they post a nice stream of questions that can provide a mental work out and that I feel can be utilised well to build problem solving amongst our students.

Today’s puzzle was this:

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It’s a nice little question. But when I use it in class I will only use the graphic, as I feel the description gives away too much of the answer. Without the description students will need to deduce that the green area is a quarter of a circle radius 80 (so area 1600pi) with the blue semicircle radius 40 (so area 800pi)  removed, leaving a green area of 800pi.

I find the fact that the area of the blue semi circle is equal to the green area is quite nice, and in feel that with a slight rephrasing the question could really make use of this relationship. Perhaps the other blue section could be removed or coloured differently and the question instead of finding the area could be find the ratio of blue area to green area.

Another option, one I may try with my further maths class on Friday,  could be to remove the other blue section and remove the side length and ask them to prove that the areas are always equal, this would provide a great bit of practice at algebraic proof.

Can you think of any further questions that could arise from this? I’d love to hear them!

This post was cross-posted to the blog Betterqs here.

Problem solving triangles

December 7, 2015 Leave a comment

Brilliant – a lovely puzzle app and a source of many little puzzlers if you follow their Facebook page. The other day, I came across this one:

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It looked like it might be interesting so I screen shot it and thought, “I’ll have a go at that later, when I’ve got a pen. It’s bound to be nice using a bit of trigonometry and angle reasoning.”

But as I thought about it I realised I didn’t need paper. The hypotenuse of the large triangle is easy enough to find (6rt2) using Pythagoras’s Theorem. You can deduce the size of the green square is then 2rt2 as the big triangle is isosceles meaning the angles are 90, 45 and 45, as the square is only right angles then the little blue triangles in the 45 degree corners must also be isosceles. Thus the two blue and the green segments of the hypotenuse are equal.

The area of the square is then way to find (8) by squaring 2rt2. A nice easy puzzle.

My first thought had been that it would take a bit of working out, but it didn’t,  it was a very straightforward question once I got going. That got me thinking, problem solving is something that I would love my students to get better at and I’m hoping to launch a puzzle of the month in January. This sort of puzzle is ideal. It will require then to build their perseverance skills as well as their problem solving skills and will give them a mental workout. I’m going to use this as a starter this week to warm them up.

This post was cross posted to the BetterQs blog here.

A latitude and longitude question

November 17, 2015 Leave a comment

A friend of mine, a geography teacher, tweeted me this question earlier:

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A lovely set of questions that I thought I’d run through here. I hope you’ve had a go at them before you read on….

The first one appeared at first to be a very simple question, just looking at the proportion of the circumference you have travelled. I then noticed the sleight of hand with the units and realised it involved an extra step. Still relatively straightforward though. Convert the distance from miles to km and work out the proportion of the circumference you have travelled. Multiply that by 36o and you have your angle, in this case 50.57 degrees to 2dp.

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The second question is more tricky. The first thing we need to work out is the radius of the circle we would get if we cut a cross section parallel to the equator 30 degrees north. In order to do this I drew a diagram  (always imperative! ).

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This allowed me to find a right angled triangle with the hypotenuse being the radius of the earth. This allowed me to find the radius of the circle I was after.

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Once I had this I could work out the distance travelled easy enough, as we have travelled 180 degrees so half the circumference.

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Or 15349 km (to the nearest km.) Here I’m assuming we want the distance we have travelled around the earth, rather than the displacement which would obviously be the diameter of the cross section or 9772km.

Area Puzzle 21

September 23, 2015 2 comments

I came across the following area puzzle on Ed’s (@solvemymaths) site.

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I found it quite an interesting idea and had a little go at solving it.

First I considered an equilateral triangle, all the angles are 60 degrees so we can see the area would be (x^2 sin (60))/2. As sin 60 = 3^(1/2)/2 it was easy enough to solve.

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I then realised I could have used Heron’s Formula, so did it that way too. Luckily I got the same answer:

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The square was a tough one:

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Then I considered a pentagon, I wasn’t sure how to approach it at first,  so I reverted to my favourite shape, the triangle. Split the pentagon into 5 congruent isosceles triangles and solved with trig.

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Wolfram alpha gave me the lovely exact answer:

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I then considered the hexagonal case, which is really just 6 equilateral triangles.

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I approached the decagon in the same manner I had the Pentagon.

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And got an equally lovely exact answer:

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I enjoyed working through these, and thought it would make a nice lesson to build resilience and cognitive activation. I also thought about what else could be done. What does the sequence of side lengths generated look like? Could an equation be formed to describe it? What if we were looking at the areas of regular polygons with side length six, what would the sequence look like then? All these would make nice investigations.

Bicycle Puzzle

September 12, 2015 5 comments

This week’s puzzle from Chris Smith is a nice contextualised simultaneous equations puzzle I intend to use next week.

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It boils down to:

2X + B + T = 135
2X + 2B + 3T = 269
X + B + T = 118

Via elimination using 1 and 3 we can see that X (number of tandems) is 17.

That gives:

34 + B + T = 135
34 + 2B + 3T = 269
17 + B + T = 118

1 and 3 rearrange to the same leaving:

B + T = 101
2B + 3T = 235

Elimination gives T (number of Tricycles) to be 33 which leave B (number of Bicycles) to be 68.

A nice little problem.

Concentric Circles Area Puzzle

June 27, 2015 6 comments

This morning I saw this post from Ed Southall (@solvemymaths):

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And thought, that looks an interesting puzzle. I’ll have a little go. I think you should too, before reading any further…

Ok, so this is how I approached it. First I drew a sketch:

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I assigned the arbitrary variables r and x to the radii of the larger and smaller circles respectively and used the fact that tangents are perpendicular to right angles, and the symmetry of isosceles triangles, to construct two right angled triangles.

I wrote an expression for the required area in r and x. Used Pythagoras’s Theorem to find an expression for x in terms or r, subbed it in and got the lovely answer of 25pi.

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An interesting little puzzle, did you solve it the same way? I’d love to hear alternative solutions.