Can you solve this geometry problem for nine year olds?

In summary, a teacher asked for help with a geometry problem involving finding the angle x. After some attempts, the solution was found to be 30 degrees using trigonometry, but it was believed that the problem could be solved without trigonometry due to its simplicity. Many people tried to solve it without success, until someone suggested constructing a new line at 20 degrees inside one of the given angles, which led to the solution. It was later discovered that this problem is known as "Langley's Adventitious Angles" and has historical significance. Some people believe it may be too difficult for a nine-year-old, but others argue that it can be explained in a lesson plan. However, there was some debate about the correctness of
  • #1
andrewkirk
Science Advisor
Homework Helper
Insights Author
Gold Member
4,116
1,712
I was in a primary school class room the other day and the teacher asked me for help with this geometry problem, that he had set for his class as an extension challenge, but then realized he couldn't do.
The known angles are marked in degrees. We have to find the angle x.

I spent five minutes trying to do it on the board and then said I'd better take it away and get back to him.

I can solve the problem using trig. The answer is ##x## is 30 degrees - or at least so close to 30 degrees that my calculator rounds it to that when at maximum accuracy (it is reached by applying an arctan function to an expression involving trig functions of 20 and 30 degrees).

But I feel the problem should be solvable without trig, firstly because the answer is such a nice round number, equal to one of the angles in the triangle, and secondly because the problem was expected to be doable by a nine-year old (albeit a very smart one).

I've spent a bit of time trying to work out a construction that shows the answer to be 30 degrees using various line extensions, perpendiculars and so on, but I just keep hitting dead ends. Geometry was always one of my weaker subjects.

Does anybody out there think they can solve it without using trig?
WPS Yr 4 Triangle Problem.jpg
?
 
  • Like
Likes Charles Link, Pavel Anni and Greg Bernhardt
Mathematics news on Phys.org
  • #2
Too much for most nine-year-old people. Sum of interior angles of a triangle being 180 degrees is a start. Notice the main largest triangle. (50+30)+20+(20+angle_ACD)=180 will give measure of angle ACD. The two intersecting segments in the left part of the large triangle form two pair of vertical angles, so the left & right oriented vertical angles have equal measure. Now use this to find measures of the other two vertical angles.

With all that, I myself do not yet find a way to get to the value of x. Is any given information missing from your posting?
 
  • #3
symbolipoint said:
Too much for most nine-year-old people. Sum of interior angles of a triangle being 180 degrees is a start. Notice the main largest triangle. (50+30)+20+(20+angle_ACD)=180 will give measure of angle ACD. The two intersecting segments in the left part of the large triangle form two pair of vertical angles, so the left & right oriented vertical angles have equal measure. Now use this to find measures of the other two vertical angles.

With all that, I myself do not yet find a way to get to the value of x. Is any given information missing from your posting?
There are four unknown angles. I named them x, y, z and w. I got 4 equations with those 4 unknowns but I think they are not independent equations. So the calculator is saying 'infinite possible solutions'. Maybe I am missing some key property of the triangle.
 
  • #4
This is a well-known puzzle. Google "Langley's Adventitious Angles". The solution involves identifying isosceles triangles and constructing additional ones.
 
  • Like
Likes Pavel Anni, mfb and cnh1995
  • #5
You can work it out using angles in a triangle and external angles. No time to do a sketch today but it all came out ok. Give all the angles letters and you can work them out one at a time, if you do it in the right order.
I would say it's a bit hard for a nine year old unless they've been used to that sort of problem. I reckon I was more like 12 when we did geometry like that.
 
  • Like
Likes Dr. Courtney
  • #6
The important step is to construct a new line CF at 20 degrees inside AC, where F is on the line AD. That forms an isoceles triangle so CF is equal to CA. Triangle ACE is also isosceles so we know that CE is equal to CA. As the angle FCE is 60 degrees, EF is also equal to CA. Finally, as the triangle DFC has 40 degrees at D and at C, it is also isosceles, so DF is equal to CF and hence also equal to EF. That means the triangle DFE has equal sides DF and EF so it too is isosceles, so the angles FDE and DEF must be equal, and equal to (180-40)/2 degrees, i.e. 70 degrees, so (x + 40) = 70.
 
  • Like
Likes Charles Link, Pavel Anni, beamie564 and 7 others
  • #7
Lovely! I am seeing the teacher again tomorrow, so I am trying to commit the solution to memory so I can draw it for him on the board. I expect he'll be pleased to know that it's actually a difficult problem.

Even now knowing the solution I can't see any way to explain why one would decide to draw the line CF out of the blue. Very impressive on the part of the person that first thought of doing that - James Mercer, according to the Wikipedia article on this problem.
 
  • Like
Likes beamie564 and RJLiberator
  • #8
Where's the lesson plan... did not know it was a difficult problem of historical significance.
 
  • #9
Sorry, I had to use Paint to Illustrate my solution, I hope that it is right
 

Attachments

  • geomproblem.jpg
    geomproblem.jpg
    11.4 KB · Views: 846
  • #10
Saph said:
Sorry, I had to use Paint to Illustrate my solution, I hope that it is right
You've assumed a right-angle where the lines cross, but that's not given in the question, and in fact the angle is not a right-angle.
 
  • #11
Jonathan Scott said:
The important step is to construct a new line CF at 20 degrees inside AC, where F is on the line AD. That forms an isoceles triangle so CF is equal to CA. Triangle ACE is also isosceles so we know that CE is equal to CA. As the angle FCE is 60 degrees, EF is also equal to CA. Finally, as the triangle DFC has 40 degrees at D and at C, it is also isosceles, so DF is equal to CF and hence also equal to EF. That means the triangle DFE has equal sides DF and EF so it too is isosceles, so the angles FDE and DEF must be equal, and equal to (180-40)/2 degrees, i.e. 70 degrees, so (x + 40) = 70.
That's a nice solution. Thing always bothers me is how the original solver thought of 'that line'. My School life was full of those seemingly arbitrary steps that got you there.
 
  • #12
The origin
 
  • #13
houlahound said:
The origin
Uh?
 
  • #14
andrewkirk said:
I can't see any way to explain why one would decide to draw the line CF out of the blue

It kind of bothered me, because if I could draw a line at 20 degrees, I must have a protractor, which I could also use to measure the angle X. In the end I decided it is OK, because you don't have to actually draw the new line at 20 degrees, you just have to say it (I mean, after all, the sketch in the OP isn't really "to scale" either).
 
  • #15
gmax137 said:
if I could draw a line at 20 degrees, I must have a protractor,
That's not the way to 'do Geometry Problems". It's like Algebra and the diagram is just symbolic and the method is Analytical. Those Greeks have a lot to answer for!
 
  • #16
andrewkirk said:
Lovely! I am seeing the teacher again tomorrow, so I am trying to commit the solution to memory so I can draw it for him on the board. I expect he'll be pleased to know that it's actually a difficult problem.
That's more or less true to form for yer average teacher. They very seldom try to solve the problems they set for homework. They tend to mark, using the Answers in the back of the Book.
Nine year olds? HA!
 
  • #17
I don't remember at what age I learned the necessary amount of geometry; it was probably at age 9. I think I would have had some chance of solving this one given a bit of time, as trying extra lines is a standard approach, and after the right one has been found it's all very straightforward, but I agree it's not at all obvious why one should try that particular line.
 
  • #18
Jonathan Scott said:
I don't remember at what age I learned the necessary amount of geometry; it was probably at age 9. I think I would have had some chance of solving this one given a bit of time, as trying extra lines is a standard approach, and after the right one has been found it's all very straightforward, but I agree it's not at all obvious why one should try that particular line.
I agree, that it's not obvious.
And this strikes me as a somewhat contrived problem.
After staring at the required parameters: stacked isosceles, equilateral, isosceles triangles, with equal legs, I have a feeling that it's a kind of "Jeopardy" type game: "Here's the solution. Now tell me, how I came up with the question".

drawn.to.scale.png


ps. I'm still working on my spreadsheet that I developed last night. I think it was very late, as I threw in a "Hail Mary" assumption near the end.

bad.omic.logic.png


I was trying to develop a system, where I could change the initial parameters, and see if everything still worked out.
I'm pretty sure that my "Hail Mary" assumption boogered the whole thing up.
Anyways.
Fun problem. :smile:
 
  • #19
triangle.gif


With a more accurate drawing, more visual clues may stand out. Angles are multiples of 10.
Connected lines from CB perpendicular to AB or from AB perpendicular to CB can only form
(20,70) right triangles.
In lower image, form EF and EG. If a 70° line from G meets a 90° line from D at the same height GE, then BEG is a mirror image of FEG. Then x=70-40=30.
 
  • #20
Jonathan Scott said:
I don't remember at what age I learned the necessary amount of geometry; it was probably at age 9.
You must have been 'Gifted and Talented' if you were doing it at that age. :smile:
There is not so much Pure geometry taught these days and the Problem Solving emphasis seems to be a lot on numerical and algebraic stuff, rather than spatial. My experience would have been limited to Covering Maths lessons (age 11 - 16), which I had to do quite frequently. (It may have changed back in the past few years, I suppose). Back in the 50s, the Geometry we learned in Primary School was pretty limited - working out perimeters and areas of flags, etc..
To be fair, this problem doesn't involve any advance geometry but it does involve that step, which would be pretty demanding for almost any School kid. The problems we were given would perhaps have 'resembled' this one but not (iirc) have required that much lateral thought.
 
  • #21
sophiecentaur said:
You must have been 'Gifted and Talented' if you were doing it at that age. :smile:
True, I was ahead of my class in everything maths-related, but at age 9 I had a sympathetic teacher who taught me some geometry and trigonometry stuff ahead of the normal schedule, and we also had a visiting student teacher who taught me algebra.

Not all schools were as helpful. At age 11 or 12, at my next school, I was so far ahead that in maths classes I was given an ancient textbook and told to sit at the back and read it silently while the rest were being given their lesson. On one very memorable occasion, the class were given a extension homework assignment to prove that if you bisect the angles of a parallelogram, the four lines form a square. The next day, near the end of the lesson, the teacher asked for a volunteer to give the proof. When no-one answered, I volunteered that I could prove that it was NOT true, which caused a tense hush. I took the chalk and showed that in general it was a rectangle but not a square (I think I showed that the diagonals of the rectangle were parallel to the sides of the parallelogram, so the angle between them was not a right angle). Just as I finished, the bell sounded for the break. The teacher left the room abruptly, looking cross, without saying anything at all, and there was a general intake of breath as if I'd just done something terrible. I was quite shocked, as I'd been quite pleased with myself for spotting the solution, and hoping that I'd be congratulated for having seen through the catch question. Instead, everyone seemed stunned and avoided talking to me for a while.

Fortunately, soon after that I won a free place (largely on the strength of my maths) at Winchester College, where the excellent and flexible teaching allowed me to push forward again, especially in maths and physics.
 
  • Like
Likes OmCheeto and sophiecentaur
  • #22
Jonathan Scott said:
The teacher left the room abruptly, looking cross,
I love it. You obviously shook his confidence. Personally, I was always delighted to give a kid the glory of showing I was wrong. But, I guess that episode was a few years ago. (Not least because of the "free places" that we could get.)
I remember the parallelogram proof. But I was out of rompers by the time I learned it. lol.
 
  • Like
Likes OmCheeto
  • #23
sophiecentaur said:
(Not least because of the "free places" that we could get.)
Yes, I discovered recently that makes me a "Fleming Boy" (after the "Fleming Report" of 1944) as well as a Wykehamist. I was at Winchester 1969-73.
 
  • #24
Come on, a 4 year old kid could solve this.
Bring a 4 year old kid since I don't know how to solve it.

Groucho%20Marx.jpg
 
  • Like
Likes OmCheeto
  • #25
Of course, I'm pretty sure that the original "extension challenge" must have been a joke by whoever provided it, possibly in the hope that a teacher might know it and get the joke.
 
  • #26
andrewkirk said:
Does anybody out there think they can solve it without using trig?View attachment 108009 ?

Strictly speaking this is not a pure geometry problem, meaning Euclidean geometry where we only use a straight edge and compass to construct, and in which there is no such thing as a "degree." There is a "right angle" in Euclid, but no degrees. So it really is a trigonometry problem. Thank goodness for algebra and trigonometry, they are big advances on Euclid.
 
  • #27
If you consider that y is another angle in triangle A?D, not 90 degrees, using algebra with the 180 degrees rule, you can find out y=60 degrees. Now, you can notice that y+x=90 degrees, which means 60 degrees + x degrees = 90 degrees, therefore x = 30 degrees!
 
  • #28
I think that the point with this problem it's to make clear that's not a sudoku puzzle, in which you apply the rule of the 180º, until you get a fit in which all triangles have exactly 180º. See next figure, what happes here? That's the point!

upload_2017-9-12_13-52-56.png
 
  • #29
In the diagram of @OmCheeto of post #18, ADEC is a 4 sided polygon, with enough information available that, using the law of cosines and law of sines, it is possible to numerically compute the lengths of the other sides, along with the length of the diagonal CD, if side AC is set equal to 1. The expressions for the sides involved quadratic equations with all kinds of terms such as ## \sin(20^o) ## and ## \cos(40^o) ## etc. in their coefficients. Solving this numerically, I was able to show that ## x ## was very nearly 30 degrees, (by the law of sines), but by this technique, I was unable to prove it precisely. ## \\ ## The results of my calculations were AC=CE=1; AD=1.35; CD=1.52; and DE=.684. The solution given by @Jonathan Scott in post #6 is really far better, but this is an alternative way to work the problem. If the angles that were given were such that the construction of post #6 was not possible, then this method would be one way that would work.
 
Last edited:
  • #30
OmCheeto said:
I agree, that it's not obvious.
And this strikes me as a somewhat contrived problem.
After staring at the required parameters: stacked isosceles, equilateral, isosceles triangles, with equal legs, I have a feeling that it's a kind of "Jeopardy" type game: "Here's the solution. Now tell me, how I came up with the question".

View attachment 191761
I'm trying to get the diagram to show up so that it can be viewed without going back to the previous page of the thread. See my post #29 and the diagram here by @OmCheeto .
 
Last edited:
  • #31
Man this is an old thread.
And I can't remember now if I solved this problem.
Quite the "head scratcher", as I recall.

llober said:
See next figure, what happes here?

No.

Charles Link said:
In the diagram of @OmCheeto of post #18, ADEC is a 4 sided polygon, with enough information available that, using the law of cosines and law of sines

That's cheating, according to the OP.

andrewkirk said:
Does anybody out there think they can solve it without using trig?

hmmm... Wondering if hdroogling with quotes works on a mac?

answer: Yes. And that appears to be the only instance of a "Hail Mary" on my computer. hmmmm...
 

1. What is the best approach to solving a geometry problem for nine year olds?

The best approach to solving a geometry problem for nine year olds is to break down the problem into smaller, simpler parts. This will help them understand the problem better and make it more manageable.

2. How do you make geometry problems more engaging for nine year olds?

One way to make geometry problems more engaging for nine year olds is to use real-life examples and hands-on activities. This will help them see the practical application of geometry and make it more interesting.

3. How can you explain geometry concepts to nine year olds in an easy-to-understand way?

To explain geometry concepts to nine year olds, it is important to use simple and clear language. You can also use visual aids such as diagrams and pictures to help them visualize the concepts better.

4. What are some common mistakes that nine year olds make when solving geometry problems?

Some common mistakes that nine year olds make when solving geometry problems include misinterpreting the problem, not using the correct formulas or methods, and not paying attention to details. It is important to guide them through the problem-solving process and encourage them to double check their work.

5. How can you help a nine year old who is struggling with geometry problems?

If a nine year old is struggling with geometry problems, it is important to provide them with extra support and guidance. This can include breaking down the problem into smaller steps, providing visual aids, and offering encouragement and positive reinforcement. It is also important to identify any specific areas where the child may need more practice or help.

Similar threads

  • STEM Academic Advising
Replies
28
Views
2K
  • General Discussion
Replies
2
Views
1K
Replies
1
Views
2K
Replies
4
Views
3K
  • STEM Academic Advising
Replies
17
Views
1K
  • STEM Academic Advising
Replies
20
Views
2K
  • Sci-Fi Writing and World Building
Replies
9
Views
2K
  • STEM Academic Advising
Replies
11
Views
1K
  • STEM Academic Advising
Replies
34
Views
8K
Replies
16
Views
1K
Back
Top