I would like to start a new column of mathematical puzzles. Scientific American has one; New Scientist has one; so I hope Science 2.0 will be happy to host one!

Preamble over, here's your "started for 10".

The diagram shows five circles, each with integer radius, all touching the base of the large triangle. The four smaller circles all touch their two neighbouring circles, with the large circle touching all four. The two sides of the triangle each touch two of the circles.

Let the radii of the circles be a, b and c, such that a > b > c.

Given that c has a length of 4 units, find the area of the large triangle. You may leave it as an exact solution.

Supplementary Question

Using the relationship between a, b and c, found in the first question, we can set the value of c to any square number and this will generate values for a and b.

However, if we set c=9 the triangle that encloses the circles flips over. The horizontal line used to construct the circles is now no longer the base of the whole triangle. We can redefine the triangle: the two equal sides are still the tangents of circles A and B, but the base is now the horizontal tangent to A so that the triangle still encloses the five circles.

What are the values of a and b for c=9 and why does the triangle flip over?

What happens for higher values of c, such as c=16, 25, 36 and so on?

Do the areas of the sequence of triangles generated by setting c as a sequence of squares tend towards a fixed vaue, or continue to increase indefinitely?

Full solution will be posted when a new question is published. There are, already, solutions below.

Update 7 March: have edited the above question for clarity and to include the extension problems discussed in the comments. Thanks to all participants.

11 March: a solution has now been published.

Circles Stuck in a Triangle

## Comments

Never mind "from". I would quite like to know how you get *to* that expression.From(a + b)^{2}= (a - b)^{2}+ [2(b*c)^{0.5}+ 2(a*c)^{0.5}]^{2}

derek_potter (not verified) | 03/03/13 | 18:26 PM

Well it wasn't entirely obvious you were talking about geometry :)

Okay, I guess you mean this:

Don't worry about me spoiling your "leave the geometry as an exercise for the reader" didactic: by far the most important step is the realization that you only need to plod through a few cases before the tangents fail to converge to the required triangle. Thanks for the neat explanation.

Umm, here's a thought. How would this puzzle go in [1 -1] space? :)

derek_potter (not verified) | 03/04/13 | 07:15 AM

Well it wasn't entirely obvious you were talking about geometryIf you hadn’t got there first, the formulae would have reminded me of

**Heron**— the 1st century Greek geometer who lived in Alexandria, not the one pictured below!

Another thought — this looks like a

**Diophantine**problem, named after an even later Greek.

—

Robert H. Olley
/ Quondam Physics Department
/ University of Reading
/ England
Robert H Olley | 03/04/13 | 12:47 PM

Well it looks to me as though c = 9, b = 18 and a = 72. As these are only the radii of the 3 circles then their diameters are double that, ie 18, 36 and 144 . As the area of a triangle is height x base divided by 2, the area of this triangle appears to be 324 X 216 divided by 2 which equals 34992 square units.

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Helen Barratt | 03/03/13 | 13:03 PM

Well it looks to me as though c = 9, b = 18 and a = 72.Why?

derek_potter (not verified) | 03/03/13 | 17:50 PM

9 = (a*b)/((sqrt(a) + sqrt(b))^2)

this yields a solution of the circles being tangent to each other and the horizontal line if a=144, b=16 and c=9, however at these radius values the tangent lines only touching the A and B circles will not intersect at the top and the three lines will not form a triangle which encloses the circles.

this yields a solution of the circles being tangent to each other and the horizontal line if a=144, b=16 and c=9, however at these radius values the tangent lines only touching the A and B circles will not intersect at the top and the three lines will not form a triangle which encloses the circles.

Rick J (not verified) | 03/04/13 | 11:11 AM

Richard Mankiewicz | 03/04/13 | 21:35 PM

Sorry, connection here is flakey; having loaded the image here's the text.

The original image is, in fact, to scale for the solution c=4, with then b=9, a=16. But changing the numbers has changed the geometry - something that the numbers alone don't tell you! Not immediately, anyway.

I have left in the new diagram the original horizontal used in constructing the circles. Notice that the smallest circle is not involved in the construction of the triangle; it is needed to calculate the other two radii.

Most diagrams in competition and exam questions always have the caveat "diagram not to scale". In this case it should have read "ignore the diagram"!

I should then change the definition of the triangle to "an isosceles triangle bounded by the common tangents of the two larger circles and the horizontal line touching the large circle." That seems to cover both solutions.

I think I will put the two questions together (c=4 and c=9) for my pre-Olympiad students!

Thanks also for the great diagrams about the first part of the question. Link together the three centres of the circles and either use the sums of areas or the distances along the horizontal; the latter is much faster.

The original image is, in fact, to scale for the solution c=4, with then b=9, a=16. But changing the numbers has changed the geometry - something that the numbers alone don't tell you! Not immediately, anyway.

I have left in the new diagram the original horizontal used in constructing the circles. Notice that the smallest circle is not involved in the construction of the triangle; it is needed to calculate the other two radii.

Most diagrams in competition and exam questions always have the caveat "diagram not to scale". In this case it should have read "ignore the diagram"!

I should then change the definition of the triangle to "an isosceles triangle bounded by the common tangents of the two larger circles and the horizontal line touching the large circle." That seems to cover both solutions.

I think I will put the two questions together (c=4 and c=9) for my pre-Olympiad students!

Thanks also for the great diagrams about the first part of the question. Link together the three centres of the circles and either use the sums of areas or the distances along the horizontal; the latter is much faster.

Richard Mankiewicz | 03/04/13 | 22:02 PM

Using my "ice cream cone" diagram, this is the way I found the area; there are undoubtedly others.

The line joining the centres of the large (L) and medium (M) circles bisects the angle at the right-bottom of the trapezium. Dropping a perpendicular from the centre of L to the horizontal construction line, the angle between this and the bisector is A, such that sinA=3/5 - use the ditances found in the first part of the question. We can then calculate the distance from that perpendicular to the bisected angle; this comes to 108. The angle B at the top-right corner of the trapezium is such that tanB=24/7. Finally we can use similar triangles and the diameter of L to calculate the required base and height of the whole large triangle.

I get the answer 884736/7 or 126390 + 6/7

The case where c=4 also has a denominator of 7 in the final answer.

If anybody wishes to try the case c=16, the method is fairly quick once it has been established.

Half this question comes from a pre-Olympiad maths paper - I just ended up complicating it! I think it now makes an even better question, with the triangle flipping between c=4 and c=9.

Thanks to everyone who took part and for the invaluable feedback.

The line joining the centres of the large (L) and medium (M) circles bisects the angle at the right-bottom of the trapezium. Dropping a perpendicular from the centre of L to the horizontal construction line, the angle between this and the bisector is A, such that sinA=3/5 - use the ditances found in the first part of the question. We can then calculate the distance from that perpendicular to the bisected angle; this comes to 108. The angle B at the top-right corner of the trapezium is such that tanB=24/7. Finally we can use similar triangles and the diameter of L to calculate the required base and height of the whole large triangle.

I get the answer 884736/7 or 126390 + 6/7

The case where c=4 also has a denominator of 7 in the final answer.

If anybody wishes to try the case c=16, the method is fairly quick once it has been established.

Half this question comes from a pre-Olympiad maths paper - I just ended up complicating it! I think it now makes an even better question, with the triangle flipping between c=4 and c=9.

Thanks to everyone who took part and for the invaluable feedback.

Richard Mankiewicz | 03/05/13 | 02:21 AM

Well I'm afraid I don't understand how your answer can be right, if you think the area of that large triangle is 126390 + 6/7 simply because even at a kindergarten level if the radius c = 9 (so its diameter d is 18) and the area of any triangle is height x base divided by 2, then the area of this triangle in your diagram appears to be about height 342 (or roughly d x 19) x base 216 (or roughly d x 12) divided by 2 which equals 36936 square units.

Above is a simplistic, very rough diagram to illustrate these guesstimates with the smallest circles as the measuring units, which still must give a rough approximation of the area of your large triangle.

.

Above is a simplistic, very rough diagram to illustrate these guesstimates with the smallest circles as the measuring units, which still must give a rough approximation of the area of your large triangle.

.

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Helen Barratt | 03/05/13 | 06:10 AM

Helen

it has been discussed above that the diagram is wrong and I posted a new one within the comments and - just now - have also added it to the original post.

Also, please check your solutions for c=9. Once you have the right numbers and try to reconstruct the geometry, you will see that the large circle is much bigger.

it has been discussed above that the diagram is wrong and I posted a new one within the comments and - just now - have also added it to the original post.

Also, please check your solutions for c=9. Once you have the right numbers and try to reconstruct the geometry, you will see that the large circle is much bigger.

Richard Mankiewicz | 03/05/13 | 11:44 AM

How can the original diagram be wrong? We can see with our own eyes that its just a triangle containing 3 circles of different sizes that touch eachother and the triangle at the different points that you mentioned.

The only circle that we know the measurement for, is the smallest circle and that has a radius of 9 units. How can any of that be wrong so far, when we can see it all clearly in the diagram? Surely units means nothing really, the c radius could be 9 turtles feet but I still don't see why the smallest circle's radius measuring 9 turtles feet could in any way be wrong or why we can't then calculate the area of the triangle in the now 'wrong' diagram just from that?

I suppose that what you are saying is that your claim that the other circles' all have integer turtle feet radii values of b and a is what was really wrong and that they couldn't all be integers, is that right? Surely that depends on what the integer value of the unit or the unknown turtle feet value really is?

Even if the values of the radii being integers is somehow wrong, by using the smallest circle's diameter of 18 units (or say one small turtle circle with a diameter 18 units or turtle feet) as the measuring unit for the triangle, we can see that the triangle in the wrong diagram is roughly 19 small circles or turtles high and 12 small circles or turtles wide, therefore the 'wrong diagram' shows a triangle with an approximate height of 342 (18 units x 19) x base 216 (18 units x 12) divided by 2 which equals 36936 square units or turtles feet which equals 2052 square turtles. I know, turtles aren't square they're round!

So please can you confirm that the diagram was not really wrong, it was your stipulation that the other circles b and a radii were both corresponding integer values to radius c that was wrong and if so, how do we know that when we don't even really know the value of the unit?

The only circle that we know the measurement for, is the smallest circle and that has a radius of 9 units. How can any of that be wrong so far, when we can see it all clearly in the diagram? Surely units means nothing really, the c radius could be 9 turtles feet but I still don't see why the smallest circle's radius measuring 9 turtles feet could in any way be wrong or why we can't then calculate the area of the triangle in the now 'wrong' diagram just from that?

I suppose that what you are saying is that your claim that the other circles' all have integer turtle feet radii values of b and a is what was really wrong and that they couldn't all be integers, is that right? Surely that depends on what the integer value of the unit or the unknown turtle feet value really is?

Even if the values of the radii being integers is somehow wrong, by using the smallest circle's diameter of 18 units (or say one small turtle circle with a diameter 18 units or turtle feet) as the measuring unit for the triangle, we can see that the triangle in the wrong diagram is roughly 19 small circles or turtles high and 12 small circles or turtles wide, therefore the 'wrong diagram' shows a triangle with an approximate height of 342 (18 units x 19) x base 216 (18 units x 12) divided by 2 which equals 36936 square units or turtles feet which equals 2052 square turtles. I know, turtles aren't square they're round!

So please can you confirm that the diagram was not really wrong, it was your stipulation that the other circles b and a radii were both corresponding integer values to radius c that was wrong and if so, how do we know that when we don't even really know the value of the unit?

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Helen Barratt | 03/05/13 | 16:22 PM

Looking at your previous comments, your solutions for a and b are wrong. There are full explanations in the other comments and Sascha has created a very good post too.

One thing that has not been explicitly stated is that there is a very neat relationship between the 3 radii. Using the diagrams above, creating trapeziums from the 3 centres of the circles, and a bit of algebra we come to:

insert c=9 and we get a problem in unit fractions. There is only one solution, so a and b can be calculated.

One thing that has not been explicitly stated is that there is a very neat relationship between the 3 radii. Using the diagrams above, creating trapeziums from the 3 centres of the circles, and a bit of algebra we come to:

insert c=9 and we get a problem in unit fractions. There is only one solution, so a and b can be calculated.

Richard Mankiewicz | 03/05/13 | 21:20 PM

Richard, you obviously didn't understand much of what I was saying or asking. Also I don't have a solution for a or b and never claimed to have one, I just said that they looked like they were 18 and 72! And just in case you think I am some sort of mathematical dunce, you are wrong, I have always been very good at maths. Also, as far as I can see the ultimate question is to calculate the area of the triangle in the diagram and so far no one has managed to do this correctly, or am I wrong there too? I'm just looking at the problem differently to you, in much more simplistic terms ;-

1. There was a diagram that showed 3 different types of circles stuck in a triangle. Nothing wrong with that so far.

2. The smallest circle had radii of 9 unknown units, nothing wrong with that either.

3. The circles touched each other in a particular way that was shown in the diagram (so that was visibly correct and not wrong).

4. You claimed that all of the circle radii were integers. Maybe that was wrong? How do you know?

5. You asked us to calculate the area of the triangle. Nothing wrong with that, its still possible.

All that I am saying is that it is still possible to roughly calculate an approximate or even an accurate area of the triangle, by simply using the smallest circle as a measuring unit, are you saying that I am wrong?

I am also questioning exactly how and why you think and claim to know that the first diagram was wrong? I asked if it was because you said that all of the radii had integer values? Are you saying it was wrong simply because Sascha's formulae to calculate both the area and the values of a and b didn't work? I then questioned how you can know the integer values of any unit that has no value? If you still can't answer my questions then can you at least just confirm that it is because you either do or do not understand my questions and next time please don't tell me that I have made any claims that I haven't actually made.

1. There was a diagram that showed 3 different types of circles stuck in a triangle. Nothing wrong with that so far.

2. The smallest circle had radii of 9 unknown units, nothing wrong with that either.

3. The circles touched each other in a particular way that was shown in the diagram (so that was visibly correct and not wrong).

4. You claimed that all of the circle radii were integers. Maybe that was wrong? How do you know?

5. You asked us to calculate the area of the triangle. Nothing wrong with that, its still possible.

All that I am saying is that it is still possible to roughly calculate an approximate or even an accurate area of the triangle, by simply using the smallest circle as a measuring unit, are you saying that I am wrong?

I am also questioning exactly how and why you think and claim to know that the first diagram was wrong? I asked if it was because you said that all of the radii had integer values? Are you saying it was wrong simply because Sascha's formulae to calculate both the area and the values of a and b didn't work? I then questioned how you can know the integer values of any unit that has no value? If you still can't answer my questions then can you at least just confirm that it is because you either do or do not understand my questions and next time please don't tell me that I have made any claims that I haven't actually made.

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Helen Barratt | 03/05/13 | 22:01 PM

Gerhard Adam | 03/05/13 | 22:34 PM

... simply using the smallest circle as a measuring unit, are you saying that I am wrong?

Yes, because you are presuming that the image is drawn exactly to scale and that the individual components can be used for direct measurement.

Gerhard, what scale are you talking about? It doesn't matter what scale exists in anyone's head all I am saying is that the diagram exists in its own right and therefore the area of the triangle in the 'wrong' diagram can be calculated simply by using the smallest circle's radii of 9 units as an approximate measuring unit to calculate the approximate height and base of the triangle, and as I said above :-

' I don't understand how Richard's answer can be even vaguely right, if he thinks that the area of that large triangle is 126390 + 6/7 simply because even at a kindergarten level if the radius c = 9 (so its diameter d is 18) and the area of any triangle is height x base divided by 2, then the area of this triangle in your diagram appears to be about height 342 (or roughly d x 19) x base 216 (or roughly d x 12) divided by 2 which equals 36936 square units'.

—

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Helen Barratt | 03/06/13 | 00:17 AM

Rereading all this at leisure, I can now see that no one except for me has even hazarded a guess at the area of the triangle in the 'wrong diagram' and that Richard's calculation of the area of the large triangle was for his new, 'ice cream cone' diagram. Forgive me for not initially realising that ice creams can only exist in some people's heads here if they are upright and that people here are constantly running around altering the goal posts, sorry I mean the diagram, creating a new 'correct' diagram because their calculations for the 'wrong' diagram triangle's area don't work!

Now, can someone please answer my questions above about why the first diagram was wrong and why it was not possible to calculate the area of the triangle in the first wrong diagram? Are you all going to tell me that the first diagram has magical properties that mean that the large triangle with its trapped circles, that any idiot can see really does exist, regardless of any scale, cannot be calculated?

Now, can someone please answer my questions above about why the first diagram was wrong and why it was not possible to calculate the area of the triangle in the first wrong diagram? Are you all going to tell me that the first diagram has magical properties that mean that the large triangle with its trapped circles, that any idiot can see really does exist, regardless of any scale, cannot be calculated?

—

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Helen Barratt | 03/06/13 | 08:44 AM

The first diagram is correct for c=4 and gives b=9 and a=36. I assumed the next solution was similar, with a steeper triangle - I was wrong.

But that wasn't the question in this case - I set c=9, the next whole number solution, which gives b=16 and a=144. These are the solutions for the problem that I described. However, if you then try to calculate the area from the diagram using the numbers from the solution, you will find that it doesn't work - the angle of the right-hand corner of the triangle is greater than 90 degrees!

Trying to estimate the area using the c=9 circle but the wrong diagram will result in the wrong area. If you try your estimate using the same method but using the ice-cream cone diagram you should get something at least the same order of magnitude. The diagram is to scale.

Sascha has also added a small diagram for the next solution, c=16, b=25, a=400. Indeed, if we analyse the sequence of solutions, it is the very first one that is anomalous - all the rest will form a triangle pointing downwards. c=1 is not a valid solution as we required that a>b but in this case the only whole number solution is a=b=4.

But that wasn't the question in this case - I set c=9, the next whole number solution, which gives b=16 and a=144. These are the solutions for the problem that I described. However, if you then try to calculate the area from the diagram using the numbers from the solution, you will find that it doesn't work - the angle of the right-hand corner of the triangle is greater than 90 degrees!

Trying to estimate the area using the c=9 circle but the wrong diagram will result in the wrong area. If you try your estimate using the same method but using the ice-cream cone diagram you should get something at least the same order of magnitude. The diagram is to scale.

Sascha has also added a small diagram for the next solution, c=16, b=25, a=400. Indeed, if we analyse the sequence of solutions, it is the very first one that is anomalous - all the rest will form a triangle pointing downwards. c=1 is not a valid solution as we required that a>b but in this case the only whole number solution is a=b=4.

Richard Mankiewicz | 03/06/13 | 09:05 AM

Thanks for explaining how the solutions came about Richard. So the problem with no one being able to calculate the area of the triangle in the 'wrong' diagram, resulted from your stipulation that the circle radii must all be integer values as I had suspected. You said that the radius of the smallest circle c is 9 units and that a > b > c , all of which can still be happily represented by the 'wrong' diagram. So you could theoretically still calculate the area of the larger triangle in the 'wrong' diagram by simply using the smallest circle's diameter of 18 units as a measuring stick, as I originally suggested, without any need to know the unit or final real values of b and a as long as they really were integer values and you are saying that they can't be if c = 9.

This brings me back to my other question that surely the unknown value of the unit or possible hypothetical 'turtle foot' must also be relevant here? For example, if the value of a 'unit' or 'turtle foot' in my example above, is .5 then this halves your calculated integer radii values and if they were calculated as being odd (ie not even) integer unit values, as in c = 9 units, then they are no longer really an integer radius value once this unit value has been applied are they? You said :-

You said above that 'The first diagram is correct for c=4 and gives b=9 and a=36.'

So what is the correct area of the triangle in the first 'wrong' diagram with these new, circle radii, integer unit values?

This brings me back to my other question that surely the unknown value of the unit or possible hypothetical 'turtle foot' must also be relevant here? For example, if the value of a 'unit' or 'turtle foot' in my example above, is .5 then this halves your calculated integer radii values and if they were calculated as being odd (ie not even) integer unit values, as in c = 9 units, then they are no longer really an integer radius value once this unit value has been applied are they? You said :-

The diagram shows five circles, each with integer radius, all touching the base of the large triangle. The four smaller circles all touch their two neighbouring circles, with the large circle touching all four. The two sides of the triangle each touch two of the circles.

Let the radii of the circles be a, b and c, such that a > b > c.Maybe you should have said five circles, each with an integer unit radius, because until you know the unit value you can't know if the circle really has an integer radius can you?

Given that c has a length of 9 units, find the area of the large triangle.

You said above that 'The first diagram is correct for c=4 and gives b=9 and a=36.'

So what is the correct area of the triangle in the first 'wrong' diagram with these new, circle radii, integer unit values?

—

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Helen Barratt | 03/06/13 | 11:25 AM

So the problem with no one being able to calculate the area of the triangle in the 'wrong' diagram, resulted from your stipulation that the circle radii must all be integer values as I had suspected.There cannot be a "problem" with stipulating that a, b and c are integers. Remove that and what is currently a puzzle requiring a little bit of mental agility degenerates into a boring exercise in school geometry.

derek_potter (not verified) | 03/06/13 | 11:52 AM

Actually the problem was that no one could calculate the area because the radii were integers

BTW I thought that your diagrams above were great! So do you know what the area of the big triangle in the 'wrong' diagram with c = 4 units is now or even the area of the big triangle in the new diagram with the circles stuck in the trapezium in which c = 16?

**and**Richard stipulated that c must equal 9 units. If he changes it to c must equal 4 units and all the radii must be integer unit values, then it is still a very interesting problem and the circles are still stuck in a triangle and not a trapezium.BTW I thought that your diagrams above were great! So do you know what the area of the big triangle in the 'wrong' diagram with c = 4 units is now or even the area of the big triangle in the new diagram with the circles stuck in the trapezium in which c = 16?

—

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Helen Barratt | 03/06/13 | 12:07 PM

Yes, but that isn't what you said.

And no, I haven't bothered to work out the area. I expect there is some neat bit of geometry to use but I haven't got the time to think about it.

And no, I haven't bothered to work out the area. I expect there is some neat bit of geometry to use but I haven't got the time to think about it.

derek_potter (not verified) | 03/06/13 | 12:10 PM

Didn't I, I thought I did? Well in case everyone has forgotten, the problem was to calculate the area and no one has done this correctly yet.

—

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Helen Barratt | 03/06/13 | 12:15 PM

Richard Mankiewicz | 03/06/13 | 09:15 AM

The diagram shows five circles, each with integer radius, all touching the base of the large triangle. The four smaller circles all touch their two neighbouring circles, with the large circle touching all four. The two sides of the triangle each touch two of the circles.

Let the radii of the circles be a, b and c, such that a > b > c.You said 'Thanks, and all further solutions will look similar' but how can this be a solution if it doesn't solve the criteria above defined in the problem? Surely we never solved it because it was an unsolvable problem? So now you have radically changed the problem and the criteria to fit a non solution!

Given that c has a length of 9 units, find the area of the large triangle.

Why not just make c = 4 and all circles have integer unit radii then everything will be fine and you can stay with the original diagram which is much more interesting than your ice cream cone and now this latest one that Sascha has generated, where the 3 circles don't even all touch the base line of the triangle. Also, if you don't stick with the original 'wrong' diagram you will have to change the name of this blog to 'circles stuck in a trapezium' ;)

—

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Helen Barratt | 03/06/13 | 11:42 AM

I must admit that I really don't understand your problem. "The question" is NOT the diagram - "the question" is the text. If I had written the text and had no picture whatsoever, how would you solve it? That's what others have done - they followed the instructions in the text - they found the integer solutions (THAT isn't a problem either) - then using those values to calculate the area found that the original diagram is the wrong way round.

OK, I'll rewrite the original question - split it into two parts. back in an hour - taking daughter to school!

OK, I'll rewrite the original question - split it into two parts. back in an hour - taking daughter to school!

Richard Mankiewicz | 03/06/13 | 18:30 PM

Well done! Now someone just needs to calculate the area of the large triangle, ha ha! I will try later........

—

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Helen Barratt | 03/06/13 | 19:13 PM

"The question" is NOT the diagram - "the question" is the text. If I had written the text and had no picture whatsoever, how would you solve it?Well in some ways that would have been a lot better because Sascha and your solutions don't fit the 'question' in the text. Your text is describing a hypothetical picture of a diagram in which :=

'... five circles,So there would have been no picture and no solution, end of story. Sascha's diagram would be wrong and so would your ice cream cone diagrameach with integer radius,all touching the base of the large triangle. The four smaller circles all touch their two neighbouring circles, with the large circle touching all four. The two sides of the triangle each touch two of the circles. Let the radii of the circles be a, b and c, such that a > b > c.

Given thatc has a length of 9 units, find thearea of the large triangle.You may leave it as an exact solution.

**because the five circles don't all touch the base of the large triangle and c does not equal 9**. The only diagram that satisfies all of the criteria except for c = 9 units is the original 'wrong' diagram. Change c to 4 in the text and then the original diagram will be correct and everything will be OK.

—

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Helen Barratt | 03/06/13 | 20:09 PM

A solution has now been published...

http://www.science20.com/florilegium/blog/solution_five_circles_stuck_tr...

http://www.science20.com/florilegium/blog/solution_five_circles_stuck_tr...

Richard Mankiewicz | 03/10/13 | 23:38 PM

^{2}= (a - b)^{2}+ [2(b*c)^{0.5}+ 2(a*c)^{0.5}]^{2}it is suggested to use squares for c (as you did) and b in order to make a integer:

a = c*b > b > c

Eg:

36 > 9 > 4

144 > 16 > 9Now the area is straight forward.