## Finding the angle problem

So today I wanted to look at this problem that I saw on a friends Facebook page:

It struck me as quite interesting so I had a think about it. I had no pen or paper with me but did some working out and got an answer. When I sent it to him he said he had got something different so we both set about trying to work out who had made the error and where that error was, and even at one point postulated that perhaps there were multiple solutions. In the end, it turned out that I had made an incorrect inference partway through my working. This lead to a lot if great discussion and thinking about why this was wrong and I’m hoping to write a longer post to discuss that later.

My initial incorrect answer was that alpha was 30.

I should have noticed straight away that this couldn’t be right, if we look at a sketch:

Seen it? How about now:

If alpha was 39, then angles DCE and CED would both need to be the same as they are base angles in an isoceles, but ECD would be 90 so if they were both 90 the lines would be parallel and not contained within the same triangle! This highlights to me the importance of completing the good old “common sense check”.

Anyway, when trying to work out who was wrong I tried a different method and git the answer my friend had, so here’s that solution:

Start with a sketch, obviously.

I know angle CAB is alpha, and I know triangle CBA is isoceles so angle ACB is also alpha.

I know angle CBD is 2 alpha from the exterior angle theorem and that gives that CBD is also 2 alpha due to base angles in an isoceles. I also used base angles in a different triangle to get angle EBD is alpha. At this point I thought that this would mean angle BCD was alpha too, as I thought the triangles EDC and BCD must be congruent, but having already jumped to one incorrect conclusion in my earlier solution I thought I’d better justify this:

Angle BMD is 180 – 3 alpha using the angle sum of a triangle. Angle EMC is the same using vertically opposite angles (even though I seem to have written the wrong thing in my workings!) Then using the angle sum again angle ECM is 2 alpha. As its isoceles ECD must also be 2 alpha. This gives me the angle I needed, and also justifies my earlier thought on the congruent triangles.e

From here I had all 3 angles of triangle ACD in terms of alpha so could get the solution using the angle sum:

5 alpha = 180

Alpha = 36.

I very much enjoyed this problem, and in particular the maths I thought about and investigated due to making an incorrect assumption. I think that there is often more to me learned from a mistake than from the correct answer and I feel here that may be true. I’d be interested to see any other solutions to this problem and I will endeavour to write up my thoughts in the other stuff that arose soon.

## An irrational triangle

The sad news about Don Steward last week prompted many people to share blog posts and tweets about how he has helped and inspired them over the years. One of the people who blogged about him was Jo Morgan (@mathsjem) who wrote this piece. Jo had the pleasure of knowing Don personally and it was really nice to read the things she shared. If you missed it, do give it a read.Today I want to look at the problem she shared at the end of the post, which she says is her favourite Don Steward problem:It’s not one I remember seeing before, although I have spent so long on Don’s fantastic median website it is highly probably I have seen it. But I thought I’d have a go. Usually when working out an are of a triangle with 3 sides I will apply Heron’s formula, but I looked at this one and realised I would end up with a massive expansion of surds and quite possibly nested radicals and that would take a long time to work through by hand, so I considered other options. I thought a locigal first attempt would be to use trigonometry. I drew a sketch:

Labelled one if the angles x. I can then work out cos (x) fairly easily:

I did think I might get stuck at this point, with the limitation in the question prohibiting calculators, but actually it worked out really nicely. If I know cos (x) I can use a right angled triangle to work out sin (x):

Here I only have to square 1 (1) and the root of 26 (26) do a subtraction and take the square root of 25 (5) to complete my triangle using Pythagoras’s Theorem. All easily done without a calculator.Now I have sin x I can work out the area using area = absin(c)/2

When I started playing with the surds I wasn’t quite expecting it to fall out so nicely. Who’d have thought a triangle with 3 irrational side lengths would have such a nice integer area? I think I will explore this more when I get time. See how easy this type of triangle is to generate and if there is a rule to triangle of this sort occurring.

I said earlier that I didn’t think I’d seen this one before. But now I think about it, I think I may have. I think I remember giving it to a year 12 class maybe last year and some of them cheating and using the calculator but not getting the right answer as it had rounded the cosine and sine values, I will need to check this later on a calculator too. If this is the case, then I think this is a great discussion point when looking at rounding and why we shouldn’t round too early. I’m also left wondering how easy/difficult it would have been to get to the nice answer using Heron’s Formula, so I will give that a go in due course too.

## 4 squares and a rectangle

Today’s puzzle comes from Catriona Shearer (@cshearer41):

It’s one that gave me a little trouble early on but finished up with a lovely solution. Have a go before you read on.

Here’s one of a few aborted efforts:

So my solution started, as usual, with a sketch. I worked on my sketch and came up with this:

Which proved to be the key. But I realise this is not a very easy to follow solution so I’ll try to explain a bit. Here is a second diagram, this time I have included some vertex labels:

Starting with triangle ABF I labelled side FB as a and FA as b, I know the side AB is 1 aa it is a side of a square of area 1. Angle BFA is a right angle so if angle FAB is x then angle ABF IS 90 – x. As angle ABC is 90 and FG is a straight line that means angle DBG is also x (180 – 90 – (90 – x)).

This means triangles ABF and BDG are similar (both have one angle 90, one x and the other 90-x). As the hypotenuse of BDG is 2 and the hypotenuse of ABF is 1 (from the side length of the squares being 1) then we know that GB is 2b, and GD is 2a. At this point I considered reflecting DBG to get BED but decided against that root.

I worked out that angle JDI (where J is directly above I such that JI and GK are perpendicular) is x. (Again 180 – 90 – (90- x)). IJD is 90 by design and the hypotenuse DI is 2 hence triangle IJD is congruent to BGD which means IJ is 2a.

However, angle JIK is also x (HIK is 90, HIJ is 90- x so JIK is 90 – (90-x)). IJK is 90 and the hypotenuse IK is 1 so triangle IJK is congruent to triangle ABF. So IJ is b.

This means as IJ is 2a, but IJ is also b we can say that b = 2a.

Back our original diagram now. Using these similar and congruent triangle I can see that the vertical side of the rectangle is a + 2b, and the horizontal one is 3a + 2b.

Subbing b = 2a in we get the horizontal side to be 7a and the vertical side to be 5a. So the area is 35a.

Almost there. We also have a number of right angled triangles we could use for the next be I chose the top left one BGD in the other diagram.

By Pythagoras’s Theorem we can see that a^2 is 1/5.

So the area is 7.

I think this is a lovely solution to a problem I very much enjoyed wrestling with. I imagine there are plenty of other approaches that you could use to solve it. Please let me know your solution either in the comments, by email or via social media.

## Semi-circles on a hypotenuse

Today’s puzzle is from Catriona Shearer (@cshearer41). It started out when I saw this tweet:

I looked at this then put my phone down and thought about it while doing something else. Only I didn’t, I misremembered entirely what the question was. This is what I thought it was:

Find the hypotenuse.

While thinking on it I though, “hang on, if the diameter of the circle is 6 then the radius of the circle is 3 which by similar triangles means the hypotenuse is 12.”

Like this.

I then thought, “that’s a bit simple for a Catriona puzzle, I wonder what it really said.”

So I looked and realised it was an entirely different puzzle. I also noticed this tweet underneath it:

I followed that link and found the puzzle as intended:

The difference being the light green dotted line. I think perhaps my misremembering and the knowledge this dotted line makes the puzzle possible rather than impossible probably influenced my solution. Firstly I sketched it out:

I realised I could reflect triangle CBD in the line CD and the other vertex would be at point E. I could then reflect this new triangle in the line EC. As ED is the size of the radius of each semi-circle and AC is a tangent it follows that the angle BCD must be one third of the angle BCA, or 1/3 of 90, or 30. Which means the angle labelled x must be 60. Although I over complicated this slightly when I wrote it out:

I know in a right angled triangle with a 60 degree angle that the hypotenuse is always double the side adjacent to the angle. So from here I knew that it was going to be the triangle I had assumed earlier. But I wanted to show it logistically.

I considered the trapezium EFCB, I could make a smaller triangle EGB that was similar to the large one. I wrote the side lengths in, but realised I still hadn’t justified them.

I felt happier once I had justified it, although now I look at it I’m a little annoyed I used x again to mean something different.

I know that this “x” is the radius of each semi circle, and that the semi circles are congruent, hence the area is x^2 pi. Or 9pi in this case.

I enjoyed this puzzle, but I wonder if I would have taken a different approach had I not know the importance of the dotted line or had not wrongly assumed the correct diameter in the first instance. How did you approach it? I’d love to hear your solutions in the comments or on social media.

## Flummoxing triangles

Earlier on in the week I came across an interesting geometry puzzle from (@puzzleprime):

I had a quick try:

Many dead ends, I did work out that ABC is an isoceles triangle and I had the embryonic thought that I would need some more lines, but wasnt sure where they’d go. I then left it for a while. I came back to it today:

I started and realised I was mainly taking the same approach. So I thought about what other lines I could draw. At this point I spotted something. If I rotated ABC so the vertex B washes at A and CB ran along where AB is we would have 2 equal sides separated by 60 degrees. So if we joined their ends we would get an equilateral triangle:

At this point I still wasn’t sure what to do next. Then while I was looking at it I realised that, despite my terrible diagram, A’B’E and ABE are congruent triangles.

This meant that the line AE bisects the angle BAA’.

As angle BAA’ is the angle that makes up the 80 degree angle with x that means x is 70.

This was a lovely puzzle that had me thinking and flummoxed at different points. I’d love to hear of you have a different solution.

## Circles and triangles

Today’s puzzle comes from Chris B (@CyclicQuads) who sent it in response to yesterday’s blog post about circles and rectangles. Here it is:

As you can see, Chris is much better on the diagram drawing front than I am. It’s a nice little puzzle that I enjoyed solving. First, I drew my own (terrible) sketch:

Note, this was actually the second sketch as I drew the 54 as angle BDC initially.

From here there were a few items of information that were obvious. Angle EOB = x, as base angles in an isoceles are equal. Also CBO is 2x as it’s an exterior angle. I also drew a line from C to O (another radius) so know that OCB is also 2x because we have another isoceles triangle.

I know that angle EOC is double angle EDC as angles at the centre are double angles at the circumference:

So I know that angle BOC is 108 – x.

From here I have all 3 angles in triangle BOC expressed in term of x, and I know the angle sum of any triangle is 180 so I can easily form and solve an equation:

This was a lovely puzzle that I very much enjoyed solving. I’d love to hear how you solved it, please let me know in the comments or via social media!

## Angle puzzle- Congruent Circles

Today’s puzzle is another from Ed Southall (@edsouthall) and another that jumped out at me as I couldn’t see an obvious solution. It also has that word “congruent” in it again!:

As I said, I didn’t immediately spot a route to the solution, so I started drawing stuff:

At this point I saw this one:

And thought it was a rombus. I drew this on the full diagram:

Briefly took leave of my senses and decided I needed to justify that it was actually a rhombus, despite there already being more than enough justification:

The I worked out the angles:

Considered circle theorems, but nothing came of them. Considered cosine rule:

Decided that was probably unnecessary:

Drew it out again, reeling I had angle BCD and that triangle BCD was, in fact, isoceles:

Then it followed as a simple subtraction to get the final answer.

This was hardly a neat or concise solution the way I achieved it, although if I’d only done the required steps it may have been nice. How did you solve it?

## Congruent Rectangles

Today this puzzle from Ed Southall (@edsouthall) caught my interest:

I think it caught my interest initally because it uses the word “congruent”, which is a great word. Don’t know why I like it so much but I do. It just looks and sounds cool.

Anyway, after that caught my eye so did the ratio, 3:4. I figured, “Hey, this puzzle uses the word congruent and probably a 3,4,5 triangle, what’s not to love?” So on I went.

As anyone would, under the circumstances, I did a sketch:

I labelled the sides up in the ratio. Because I knew the diagonal was 5 I could then get a side of the smaller triangle. Due to alternate angles I know it’s a similar triangle SF 1/2 so I can work put the other sides too.

Then it was just a case of calculating the areas:

Finding the pink area:

And writing it as a fraction:

A nice easy solve to an interesting looking puzzle. I then realised there was another possible solution. I had assumed that the lengths were this way as in the diagram one is clearly longer than the other,.so I think this is probably what Ed intended, but I did wonder what the answer would be if I’d assumed the other way:

Another nice solution.

It then occured to me that there are actually 2 more possible solutions:

All of them nice.

It occurs to me that this would be a great question to use with a class to explore their problem solving abilities. It uses pretty easy mathematics, but draws from things that are not necessarily taught together and will require students to really think.

## All the squares

Today’s puzzle comes from the mind (and felt tip pens) of Catriona Shearer (@Cshearer41):

It’s an interesting looking puzzle with 4 squares. If you haven’t yet tried it, give it a go before reading my solution.

So, first up I did a (terrible) sketch:

I noticed that the yellow square had its opposite vertices on adjacent vertices of the large dark blue square. Which means that the side of the yellow square bisects the right angle of the dark blue square. This in turn means that I can extent the line all the way to the opposite corner of the dark blue square and bisect that one two.

I then labelled the point where the green and yellow squares meet as A and the two right most vertices of the dark blue square as B and C and drew lines in to give me a triangle ABC. As AC bisects a right angle I know that angle ACB is equal to 45 degrees. At this point I labelled the point where the green and light blue squares meet as D as I noticed that this would give angel ADB as 90 degrees which is double angle ACB.

I know that when lines are drawn from each end of a chord then the angle that they make at the centre is twice the angle that they make at the circumference. So D is the centre of the circle through ABC (the circumcircle of triangle ABC).

This led me to see that A,B and C are all equidistant from D.

I had enough information to calculate CD as it is the diagonal of a square that has area 2. So by Pythagoras’s theorem the diagonal is 2, so the radius of the circumcircle is 2. This means BD and AD, the sides of the square we are interested in, are also both 2. Hence the area of the square is 4.

*I enjoyed this puzzle. It got me thinking and I used some maths that I don’t usually use while solving puzzles like this, which made it even more fun. How did you solve it? I’d love it if you let me know in the comments or on social media.*

## Puzzling puzzle

Today’s puzzle comes from @Giin1573672 :

Here’s how I went about solving it:

First I sketched it:

Then I used some similar triangles:

This gave me:

From here I considered the quarter circle:

Used pythagoras’s theorem to get a quadratic and solved it:

Obviously we can discount 2 as an answer as that would mean the side if our triangle was 0.

Typing this up I have just realised that at no point is it given that ABCD is a square, that is an assumption I have made subconsciously. I have asked from clarity from the puzzle setter if this is a fair assumption, and I’m wondering if there is anything in the question that actually fixes the length AD, or even if the length AD matters at all in this case. I’m off to ponder this more.

Right, so I’ve pondered it some and have some further thoughts. After first trying a shed load of algebra and getting nowhere I tried looking at the case where ABCD was a 2×3 rectangle:

This gave me no real solutions. I then realised some mistakes in my general algebra and tried again, this time giving:

I then ran this through Wolfram Alpha and got this:

From the implicit plot we can see that there are actual more than one sets of solutions. Some can be discounted by the diagram, for instance we know x is smaller than 2 but greater than 0, we know y is bigger than 1 but this still gives a number of solutions. My feeling is that the question was supposed to set y=2 as this nicely yields a single solution (as the other solution had x=2 and hence doesn’t fit). Either way, it’s a nice puzzle that got me thinking about a lot of things.

*How did you solve it? Do you have any other ideas on the side length AD? Please let me know in the comments or on social me*dia.