Geometry: Similar Shapes

[DaveC426913]

How exactly is the problem "deliberately misleading"?

Are you asserting that

- that in a given geometric diagram, two lines that look pretty parallel can be assumed to be parallel?

- that, in this diagram, I can assume AD and BC are parallel simply because they look parallel?
View attachment 363325
- that this diagram is "maliciously misleading"?

I just know from decades of experience that it is hard enough to teach even the most basic facts, that one does not ever have time to waste tricking 98% of people with questions like this.

So, in this diagram, if AD and BC are not parallel, you would consider this "tricking" people, yes?

 

[DaveC426913]

You're kidding, right?
Here's from post #26, as to why the figure in the problem is deliberately misleading.

And I am taking the gist of your "deliberately misleading" criteria and applying to another case:

By your logic, it seems to me, this digram is misleading, because AD and BC appear parallel, though it will turn out they are not.

Are they also taught not to trust definitions and theorems? This seems ridiculous.

I am speechless. That is a straight up straw man. And an outrageous one at that.


An angle in a diagram is not 90 degrees unless it is labeled (or proven) 90 degrees.
Two lines in a diagram are not parallel unless they are labeled (or proven) parallel.

Agree or disagree?


BE and CD cannot be assumed to be parallel in this diagram. Agree or disagree?

 

[TensorCalculus]

Edit: I have just realised that this is Q.22 from Edexcel Maths Paper 1 November 2017. I'm off to read the examiner's report.

They almost certainly will not. GCSE and A-Level exams have been developed over many years to test the ability to apply knowledge, not the ability to spot trick questions. Any (UK) exam board that let a question like this slip through the net of pre-testing would be heavily criticised within the profession.

Yeah, they do come up... Our school gives up problems like these often to practice for maths GCSE, and I have seen problems like this on the A level as well. More so for things like Olympiad/maths challenge papers, from UKMT, but also in GCSE. Or maybe our school is just being careful and therefore trying to expose us to everything, who knows. But I have heard from older students a lot about deceptive A level maths questions.

As for the pedagogical value, it makes perfect sense and does no harm to give students this as exercise that is not graded, to teach them to watch out for misleading pictures. And that includes not fooling yourself with your own exemplary diagrams that you draw when solving a problem.

But I would not put that into a graded test.

My opinion exactly.


I agree that the question is deliberately misleading, but that doesn't mean that it's malicious. Plus, in the case where x = 2, the two lines are parallel, and the diagram is correct (at least, when worrying about which lines look parallel and which don't). If the lines clearly weren't parallel, then you could argue that it is misleading for the case where x = 2. (But I do understand that x = 2 is the more obvious solution and therefore a diagram where the lines weren't parallel might have been a better option)

 

[Mark44]

@DaveC426913 you must be very fond of that quadrilateral drawing, as you've posted it three times. I assume it's a quadrilateral inasmuch as I can plainly see four sides. Or maybe that's just an assumption on my part?

For the problem from post #1, @pbuk explained why the figure is deliberately misleading; namely, that the two segments of side AD -- AE and ED -- are drawn roughly the same length but labeled with one of them being four times as long.

 

[fresh_42]

Yeah, they do come up... Our school gives up problems like these often to practice for maths GCSE, and I have seen problems like this on the A level as well. More so for things like Olympiad/maths challenge papers, from UKMT, but also in GCSE. Or maybe our school is just being careful and therefore trying to expose us to everything, who knows. But I have heard from older students a lot about deceptive A level maths questions.

I remember several anecdotes about such kinds of questions we used to tell. At least, the mathematical one of the following did happen since I knew the person who told it:

1. Physics: This flower pot at the window is warm on its room side and cold on the side facing the window. Why? (Answer: The professor turned it around before the exam.)

2. Physics: Explain the function of fuses. (The student allegedly got away with his answer that he didn't know because his mother had always forbidden him to play with them.)

3. Overheard in an elevator: "The idiot actually asked me how many times one can differentiate the constant function." complaining that once obviously wasn't the correct answer.

Admittedly, these were oral exams, but they show that unusual questions and the invitation to think outside the box really do happen.

 

[hutchphd]

It's not a "trick question". Students are taught to not trust diagrams.

My problem with this pedagogy is that my first admonition as a teacher was to "draw a rough diagram". This was engendered by my getting answers, reported to seven sig figures that were off by 8 orders of magnitude. Not good engineering.
Supplying drawings that are deliberately skewed makes a mockery of this important advice.
I prefer a good rough drawing to an 8 orders of magnitude ****-up every time. Engineers (precomputer) used to regularly do calculations on graph paper to scale. They put in a little more margin of course.
But bad pedagogy.

 

[TensorCalculus]

You will not believe the amount of times children in my year have gone into exams, not known how to mathematically solve the problem, assume the diagram is to scale and make simplifying assumptions (e.g. oh, that looks like it is half the length let me assume it is half the length) and then gotten the question right without actually applying the maths that they were supposed to in order to get the correct answer. Misleading diagrams puts a stop to this.

 

[weirdoguy]

It is a deliberate, tried-and-true technique to teach the student a critical geometry analysis skill.

@DaveC426913 are you a teacher doing that in your practice? You know, some techniques look perfect on paper, but practically they suck completely.

Besides, I have only 17 years of teaching math (and physics) under my belt as tutor/private teacher (yes, I started when I was 17), but I'm with @Mark44 on this.

Also, as a tutor I very often work with materials that school/college teachers provided, and sometimes there is pure malice on their part. There were situations where students wrote to some "higher power" about that and I was asked to be a 'reviewer'. E.g. there were two teachers (one in high-school, the other one at my own uni, not related) who did not give maximum amount of points for a perfectly good answer, because it was not clever enough! Of course they had those very clever way of reasoning prepared, but please... These people should not teach. Fortunately "higher powers" did their job, and in both cases students got their points back.

 

[DaveC426913]

I am not actually sure what you are saying here.

My problem with this pedagogy is that my first admonition as a teacher was to "draw a rough diagram".

You're telling your students to draw a rough diagram. (Not sure how that's a problem.)

This was engendered by my getting answers, reported to seven sig figures that were off by 8 orders of magnitude. Not good engineering.

You're saying, the reason you've adopted this policy is because you were getting 7 sig dig answers from students.

Supplying drawings that are deliberately skewed makes a mockery of this important advice.

I don't follow how. The drawings are a "rough diagram", similar to what you described above.

I prefer a good rough drawing to an 8 orders of magnitude ****-up every time.

OK, good. A rough drawing cannot be used to spot 90 degree angles or parallel lines.

If two lines are supposed to be parallely in a rough drawing, they will have to be indicated with '> >', right?

So, no '> >' means not parallel.

Engineers (precomputer) used to regularly do calculations on graph paper to scale. They put in a little more margin of course.
But bad pedagogy.

Right. This is bad.

Geometry is not about accurate drawings and measurements; it is about rough diagrams, properly labeled, and never making assumptions based on a diagram.

The way I read you, we are in complete agreement.

 

[DaveC426913]

You will not believe the amount of times children in my year have gone into exams, not known how to mathematically solve the problem, assume the diagram is to scale and make simplifying assumptions (e.g. oh, that looks like it is half the length let me assume it is half the length) and then gotten the question right without actually applying the maths that they were supposed to in order to get the correct answer. Misleading diagrams puts a stop to this.

Thank you. I would have thought this was plainly apparent to all.

 

[TensorCalculus]

Also, as a tutor I very often work with materials that school/college teachers provided, and sometimes there is pure malice on their part. There were situations where students wrote to some "higher power" about that and I was asked to be a 'reviewer'. There was two teachers (one in high-school, the other one at my own uni, not related) who did not give maximum amount of points for a perfectly good answer, because it was not clever enough! Of course they had those very clever way of reasoning prepared, but please... These people should not teach. Fortunately "higher powers" did their job, and in both cases students got their points back.

It's a shame that happens, genuinely. And I agree that those people should not teach. But in this case, it's less about how clever the solution is and more about the problem itself: even if the intent was to malicious, the fact still holds that the OP is learning a valuable lesson here, through the struggle. And in this case, in my eyes that makes it okay to give students this question.

Maybe on the actual GCSE the diagram should have been more accurate (though they should not draw their diagrams to scale, ever) but in this case, it's a good thing the diagram is the way it is. The OP is learning and regardless of the intent of the writer, it's a good thing in the end.

 

[DaveC426913]

...the fact still holds that the OP is learning a valuable lesson here, through the struggle.

... they should not draw their diagrams to scale, ever) but in this case, it's a good thing the diagram is the way it is. The OP is learning and regardless of the intent of the writer, it's a good thing in the end.

My smack is gobbed that we here are (apparently) not all unanimous in this view.

 

[renormalize]

My smack is gobbed that we here are (apparently) not all unanimous in this view.

Are you sure it wasn't your gob that was smacked?

 

[fresh_42]

I am really surprised that some seriously expect to take BE and CD as parallels and don't have any problems taking 12 and 3 being different, although they look almost equal. That simply doesn't make any sense.

 

[DaveC426913]

Not to mention that the problem explicitly says there are two solutions. If the student finds one solution and stops there, they have not understood the assignment. That's a lesson unto itself.

So, there's no trick here, it's merely that it's not a dead-easy problem; it requires the student to do some thinking.

 

[votingmachine]

Homework Statement: See screenshot in main body
Relevant Equations: I don't know if there is a generic equation
Maybe something like AB(SF)=AC ?

View attachment 363169
Why are there two possible values of x?

Here's what I've done so far.

AE(SF)=AD
12(SF)=15
SF=15/12
=5/4

AB(SF)=AC
8(5/4)=AC
AC=10

AC-AB=x
x=10-8
x=2

A hint please, as to my next step

You are assuming that there is only one way the triangles can be similar ... the drawing leads you there ... I suspect you need to evaluate the words and make your own drawings.

 

[fresh_42]

Not to mention that the problem explicitly says there are two solutions. If the student finds one solution and stops there, they have not understood the assignment. That's a lesson unto itself.

So, there's no trick here, it's merely that it's not a dead-easy problem; it requires the student to do some thinking.

... plus, there is a symbol available in geometry to show parallelity in pictures. //

 

[DaveC426913]

... plus, there is a symbol available in geometry to show parallelity in pictures. //

You mean >

 

[mathwonk]

What surprised me was that there seem to be more than 2 ways the triangles can be similar, but they yield only 2 results for the question asked. Did this surprise anyone else? Or am I wrong?

 

[fresh_42]

You mean >

I know it with two short lines, but anyway. If the sides were parallel, there would have been a possibility to mark it accordingly.

 

[fresh_42]

What surprised me was that there seem to be more than 2 ways the triangles can be similar, but they yield only 2 results for the question asked. Did this surprise anyone else? Or am I wrong?

I thought that there were just two orientations possible, and that is the reason why, but I admit, I haven't considered this in detail.

 

[TensorCalculus]

What surprised me was that there seem to be more than 2 ways the triangles can be similar, but they yield only 2 results for the question asked. Did this surprise anyone else? Or am I wrong?

The fact that $\angle BAE = \angle CAD$ eliminates the other options, right?

 

[mathwonk]

the triangles could be isosceles, and hence have self similarities. right? i.e. we seem to be making the unstated assumption that the similarity must fix angle A. It seems angle BAE could equal angle BEA, no?

 

[TensorCalculus]

Well if the triangles are isosceles, then there would still only be 2 answers for x, no?

EDIT: I need to write this down on a piece of paper. I'm pretty much just guessing based on what would make sense but I haven't actually tried solving for x in the case where the triangle is isosceles

 

[mathwonk]

yes, I believe that is what I said.

 

[DaveC426913]

View attachment 363376

I know it with two short lines, but anyway. If the sides were parallel, there would have been a possibility to mark it accordingly.

I've not encountered that notation.

Where I come from, | means equal length. To-wit:

AB and DC are parallel.
AD and BC are of equal length.

 

[TensorCalculus]

yes, I believe that is what I said.

Oh, I thought you meant that the question only asked for 2 results but there are more than two orientations. My bad (you will find that I am pretty good at misinterpreting online messages ).

I've not encountered that notation.

Where I come from, | means equal length. To-wit:

AB and DC are parallel.
AD and BC are of equal length.

View attachment 363377

Same here in Britain (and since the OP seems to be preparing for GCSE this is the notation that they would see).
Though maybe @fresh_42 means not on the diagram but when referencing the lines. E.g. I have seen the notation $AB\parallel CD$ before.

 

[mathwonk]

I admit I still find this confusing, but it seems to me that all 6 possible similarities can occur, still with only two final answers of course.
I mention this because it seems to me the question can be viewed as having 2 parts: 1) how many different ways can the triangles be similar, and then 2) how many different values for x do these give? Since the lesson is about not making unjustified assumptions, I did not want to assume the similarity must fix A.

 

[DaveC426913]

Well if the triangles are isosceles, then there would still only be 2 answers for x, no?

EDIT: I need to write this down on a piece of paper. I'm pretty much just guessing based on what would make sense but I haven't actually tried solving for x in the case where the triangle is isosceles

Since the two triangles are defined as similar, there must be two possible solutions (since they share an angle in the diagram).

If those two triangles happen to be isosceles, there may be more solutions, but we cannot say they are isosceles; we can only speculate that they might be. So, we can speculate that there might be more solutions conditionally. But this is well beyond the scope of the question.

 

[DaveC426913]

I admit I still find this confusing, but it seems to me that all 6 possible similarities can occur, still with only two final answers of course.

Do you mean could occur if all numbers were put to the diagram?

Because we can't just pretend, say, that the triangles are isosceles, let alone equilateral.

The diagram is fixed by what is labeled. If it's not labeled, it can't be assumed or even speculated.

Let me illustrate that:

BE is not equal in length to AE.
Even if BE were measured to be exactly 12cm, it is still not geometrically the same length as AE.

Geometrically, there are only two ways to determine that two lines are equal: by definition, or by proof.

The problem/diagram provides neither, therefore no valid solutions can depend on that condition.

 

[mathwonk]

I believe all 6 similarities are possible with the numbers as given. E.g. with the numbers as given, one can apparently have angles BAE, ABE, ADC, all equal. Then the "rotation" taking vertices A-->D, B-->A, E-->C, seems to be a similarity from triangle ABE, to triangle DAC. Does that seem right?

 

[hutchphd]

Geometry is not about accurate drawings and measurements; it is about rough diagrams, properly labeled, and never making assumptions based on a diagram.

I was not teaching a course in geometry, but engineering physics. Including a diagram that tries to misrepresent the discription seems , on its face, both unwise and unfair to the student.
I would not advocate teaching geometry without more rigor than "yep those look parallel to me........don't need no stinkin' postulate".
You also seem to misunderstand (or misrepresent?) some other things I said and would ask you to look at them again. I said them as well as I know how. Perhaps I should have included a drawing ?

 

[DaveC426913]

I was not teaching a course in geometry, but engineering physics. Including a diagram that tries to misrepresent the discription seems , on its face, both unwise and unfair to the student.

Totally 110% agree. Engineering is a completely different animal.

I would not advocate teaching geometry without more rigor than "yep those look parallel to me........don't need no stinkin' postulate".

Absolutely. In engineering, two lines can be parallel if they are measured to be parallel (to within some degree of tolerance).

You also seem to misunderstand (or misrepresent?) some other things I said

I know. I'm having trouble interpreting what you wrote. I did my best.

If I were to try to put my finger on why I'm having trouble, it would be because you're using passive voice a lot, which makes it difficult to tell what you are arguing:
"Thing X is done." (Is it good or bad that thing X is done? Which way do you want me to take that?)
"Do thing X." (OK, you're saying doing thing X is clearly the correct option.)

 

[hutchphd]

... plus, there is a symbol available in geometry to show parallelity in pictures. //

But using that symbol in the "official" drawing would be a lie.
It so happens that the assumption of || does lead to a valid solution: but it precludes others. Such is the usual nature of any "Ansatz" used for solution. That's the true wisdom here IMHO

 

[DaveC426913]

:furiously Googles "ansatz":

 

[Muu9]

I dislike problems like this where the figure is maliciously drawn way out of scale.

Given that there are two different values of x, it would be impossible to draw the triangles to scale in this problem, as if you drew them to scale for one value of x, it would appear "maliciously" drawn way out of scale for the other value of x.

 

[TensorCalculus]

Given that there are two different values of x, it would be impossible to draw the triangles to scale in this problem, as if you drew them to scale for one value of x, it would appear "maliciously" drawn way out of scale for the other value of x.

exactly! If you draw the diagram to present the two lines as not parallel, then it's out of scale for the solution x=2 where the lines would be parallel!

 

[fresh_42]

But using that symbol in the "official" drawing would be a lie.
It so happens that the assumption of || does lead to a valid solution: but it precludes others. Such is the usual nature of any "Ansatz" used for solution. That's the true wisdom here IMHO

I think this question is not about the intercept theorems, but rather about correctly analyzing the set of given conditions without adding assumptions that have not been made. To me, it is a lesson of not accepting only the obvious and thinking about the not-so-obvious. As I said, criminal investigators, but also scientists, and engineers have to deal with such situations on a daily basis. Where else should they learn this if not in their class that should teach thinking? Mathematics has been for a long time what we call a Geisteswissenschaft, and I think for a reason.

But I admit, I am against this in my mind, stubborn way of treating math classes as a place to learn algorithms instead of real math. If teachers were right, we wouldn't ultimately need any math classes any longer, since WA can do what kids learn at school; a depressing perspective.

 

[paulb203]

Thanks a lot guys.

I got there in the end, thanks to all of you, and a bit of extra help from Maths is Fun.

(Flip AEB on its head, as it were, to put E at the top, with A still at the bottom left, etc.

Other value of x = 14.5)

I found this one very tough, but then it was Higher tier (GCSE, UK), so it might be aimed at students who are hoping for an 8 or a 9. I put a lot of work in to get a 7 last year.

I’m now trying the A Level material (and revising GCSE as I go), but this was one of those questions that made me feel like giving up.

Onwards and upwards though.

Cheers.

P.S. I now know that congruent shapes are a special case of similar shapes - I think?! My brain is frazzled!
P.P.S. I also found out recently that a circle is a special kind of ellipse :)

 

[TensorCalculus]

Thanks a lot guys.

I got there in the end, thanks to all of you, and a bit of extra help from Maths is Fun.

(Flip AEB on its head, as it were, to put E at the top, with A still at the bottom left, etc.

Other value of x = 14.5)

I found this one very tough, but then it was Higher tier (GCSE, UK), so it might be aimed at students who are hoping for an 8 or a 9. I put a lot of work in to get a 7 last year.

I’m now trying the A Level material (and revising GCSE as I go), but this was one of those questions that made me feel like giving up.

Onwards and upwards though.

Cheers.

P.S. I now know that congruent shapes are a special case of similar shapes - I think?! My brain is frazzled!
P.P.S. I also found out recently that a circle is a special kind of ellipse :)

Maybe give L2 further maths a go first. It will set you up well for A level.

 

[Steve4Physics]

Other value of x = 14.5)

Well done! I hope you drew diagrams:

I found this one very tough, but then it was Higher tier (GCSE, UK), so it might be aimed at students who are hoping for an 8 or a 9. I put a lot of work in to get a 7 last year.

For those who don’t know, the UK GCSE exam' determines the ability/attainment-level of students, most commonly at around age 16. Grades 1 to 9 are awarded with 9 being the highest.

In math’s only around 5% of students achieve grade 9 (though, sadly, this figure varies regionally across the UK).

I'd guess that this question was written with the intention to differentiate (in the non-calculus sense!) the grade 9s. So the question is written with the expectation that (some) grade 9 candidates should be able to answer in full - but the large majority of candidates should not.

 

[hutchphd]

:furiously Googles "ansatz":

it has so much more pizzazz than "guess", don't you think....?

 

[Beyond3D]

...

 

[Beyond3D]

Homework Statement: See screenshot in main body
Relevant Equations: I don't know if there is a generic equation
Maybe something like AB(SF)=AC ?

View attachment 363169
Why are there two possible values of x?

Here's what I've done so far.

AE(SF)=AD
12(SF)=15
SF=15/12
=5/4

AB(SF)=AC
8(5/4)=AC
AC=10

AC-AB=x
x=10-8
x=2

A hint please, as to my next step

Let me help, it is a pretty basic geometry problem if you know the rules behind it
Assume that the triangles are congruent (scalable to eachother) and that the ratios between the sides remain constant
$$\ AB:AE=AC:AD$$
$$\frac{8cm}{12cm}=\frac{x+8cm}{15cm}$$
$$15\cdot\frac{8cm}{12cm}=15\cdot\frac{x+8cm}{15cm}$$
$$\frac{120}{12}=x+8$$
$$\ 10 = x+8$$
$$\ x=2$$

 

[DaveC426913]

Let me help, it is a pretty basic geometry problem if you know the rules behind it
Assume that the triangles are congruent (scalable to eachother) and that the ratios between the sides remain constant
$$\ AB:AE=AC:AD$$
$$\frac{8cm}{12cm}=\frac{x+8cm}{15cm}$$
$$15\cdot\frac{8cm}{12cm}=15\cdot\frac{x+8cm}{15cm}$$
$$\frac{120}{12}=x+8$$
$$\ 10 = x+8$$
$$\ x=2$$

OK, but that's only one of the two indicated solutions.

And you've based it on this assumption:
$$\ AB:AE=AC:AD$$

 

[Beyond3D]

OK, but that's only one of the two indicated solutions.

And you've based it on this assumption:
$$\ AB:AE=AC:AD$$

Yes, and that assumption is *heavily* implied graphically.

 

[DaveC426913]

Yes, and that assumption is *heavily* implied graphically.

1. In geometry, there are only three ways of determining that two lines are parallel: definition, postulate, and proof. Diagrams are not to be measured or eyeballed or heavily implied.
2. Never mind "implication" or "assumption"; it is explicitly stated in the problem that there are two solutions. (perhaps you have not fully read the assignment?)

 

[TensorCalculus]

Yes, and that assumption is *heavily* implied graphically.

Let's not start this argument all over again

Yes, maybe the diagram is misleading in order to push students to think harder (or, to make the question writer look smarter. We don't know.). But the problem says 2 solutions, and to even further assist the student in realising what the other answer is, they have asked you to state any assumptions that you have made in your workings.
The answer that the OP was struggling with (and has now found) was via not making that assumption.

 

[Beyond3D]

1. In geometry, there are only three ways of determining that two lines are parallel: definition, postulate, and proof. Diagrams are not to be measured or eyeballed or heavily implied.

2. Never mind "implication" or "assumption"; it is explicitly stated in the problem that there are two solutions. (perhaps you have not fully read the assignment?)

I know there are two solutions, but how would you get the second?

 

[TensorCalculus]

I know there are two solutions, but how would you get the second?

Do you want me to tell you or would you rather work it out yourself?
You've assumed that $AB:AE=AC:AD$ - in the other case that isn't true