# B Can you solve this geometry problem for nine year olds?

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#### andrewkirk

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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 realised 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?
?

#### symbolipoint

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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?

#### cnh1995

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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.

#### Jonathan Scott

Gold Member
This is a well-known puzzle. Google "Langley's Adventitious Angles". The solution involves identifying isosceles triangles and constructing additional ones.

#### sophiecentaur

Gold Member
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.

#### Jonathan Scott

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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.

#### andrewkirk

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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.

#### houlahound

Where's the lesson plan.... did not know it was a difficult problem of historical significance.

#### Saph

Sorry, I had to use Paint to Illustrate my solution, I hope that it is right

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#### Jonathan Scott

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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.

#### sophiecentaur

Gold Member
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.

The origin

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#### gmax137

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).

#### sophiecentaur

Gold Member
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!!

#### sophiecentaur

Gold Member
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!

#### Jonathan Scott

Gold Member
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.

#### OmCheeto

Gold Member
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".

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.

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.

#### phyti

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.

#### sophiecentaur

Gold Member
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.
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.

#### Jonathan Scott

Gold Member
You must have been 'Gifted and Talented' if you were doing it at that age.
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 text book 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.

#### sophiecentaur

Gold Member
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.

#### Jonathan Scott

Gold Member
(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.

#### jonjacson

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.

#### Jonathan Scott

Gold Member
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.

"Can you solve this geometry problem for nine year olds?"

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