AlefBet
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Are the interior lines perpendicular to the sides, or do they bisect the angles, or neither?

Edit: Hmm. They can't be perpendicular. The areas don't add up for that.

Edit: Er, that is, I mean they can't bisect. They don't add up. Hmm. Maybe I should delete this whole post. Nah, I'll leave it as a monument to hastiness and lack of double-checking.

[Edited by AlefBet at Local Time:03-03-2005 at 04:07 PM]

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eytanz
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RoboBob3000 wrote:
I think positivity is a restriction, because then there'd be no end to negative numbers containing the word "thousand".

As opposed to the finite amount of positive numbers containing the word "thousand"?

[Edited by eytanz at Local Time:03-03-2005 at 05:05 PM]

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Fried Chips
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eytanz wrote:
As opposed to the finite amount of positive numbers containing the word "thousand"?

But there is a lowest positive number containing the word "thousand" (1000, obviously), whereas there is no lowest negative number ...

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stigant
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you may make no presumptions about any of the angles or the lengths of any of the sides. You know only the areas.

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DiMono
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Is it drawn to scale?

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stigant
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It might be. But I never make any promises about that sort of thing. Which is to say, no its not drawn to scale.

Let me give you guys a bit of a hint. There are lots of different ways that this triangle might look, and still have the areas indicated. It might be very wide, in which case it would be very short. Or it might be very skinny, and therefore very tall. Or one of the angles might be 90 degrees, or 89 degrees, or 88 degreees, or.... However, in all of those configurations, the area of the whole triangle is the same.

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DiMono
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That's exactly the principal I'm working off of right now.

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DiMono
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Area of a triangle is 1/2*w*h. Assuming the base is 6, that means the height of the bottom triangle is 2. Since the area of it and the 2 is 8, that means the height of the 6 and the 2 triangles combined is 16/6=2 2/3. Similarly, the height of the 6 and 3 triangle is 18/6=3. Identifying the similar angles, we get this picture:



Since for the properties I used to identify those values angles don't matter, let's pretend it's a right triangle to identify the length of one of the intersecting lines. The proportions won't change, just the type of triangle. Since the area of the 3 triangle must be 1/2*3*w, the distance from the intersection to the left side of the triangle is 2. That means the intersection is at (2,2), which means the line from the bottom-left corner is at a 45 degree angle, so it ends at (2 2/3,2 2/3). Now we get this picture:



Now we have enough information to figure out the total height of the triangle. A similar triangle to the large one is formed by the line from (0,2 2/3) to (2 2/3,2 2/3). The ratio between the total height and the height of the missing segment will be 6:2 2/3, or 18:8.

3+n:n ~ 18:8
24+8n=18n
24=10n
n=2.4

So the total height is 5.4 and the width is 6, so the total area is 6*5.4/2, or 16.2



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Fried Chips
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Is it 14.4? My answer seems too rational for a mistake...

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Fried Chips
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DiMono, I drew one such triangle to scale and my answer seems closer ... but I may be wrong

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stigant
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Fried Chips has the correct answer, but I would like to see a proof without assumptions before I pass the torch.

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DiMono
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Could you please point out the error in my calculations?

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Maurog
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File: 1.png (1.6 KB)
Downloaded 445 times.
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How about just building an example from scratch?

Let's take, say, the default triangle with area 6 (that would be the 3,4,5 triangle). Now, you know the areas of the adjoining triangles, so you know how far the lines should be extended (1 and 2 respectively). Now, to find the 3rd point of the triangle - solving two equations we get the point (7.2,4.8). The rest is simple - calculating the area of the triangle by taking the area of the trapezoid formed by the points (0,3),(0,0),(7.2,0) and (7.2,4.8), which is 28.08, and deducting the two excess triangles which had areas 7.68 and 6. 28.08-7.68-6=14.4, which must be a solution.
Assuming there is a single solution, it must be it.

[Edited by Maurog at Local Time:03-03-2005 at 07:24 PM: retroactive img link]

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Fried Chips
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Yes, but stigant wants a solution with no assumptions. I'm working on one at the moment.

[Edited by Fried Chips at Local Time:03-03-2005 at 07:30 PM]

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DiMono
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Maurog: in which case, there is an error somewhere in my calculations, and I'd like to know where it is

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Fried Chips
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DiMono, you define n as starting from the line y=8/3, yet you twice calculate the total height as 3 + n rather than 8/3 + n.

[Edited by Fried Chips at Local Time:03-03-2005 at 08:07 PM]

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stigant
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A similar triangle to the large one is formed by the line from (0,2 2/3) to (2 2/3,2 2/3)

How do you know that the top of the area 2 triangle is (2 2/3, 2 2/3)?

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DiMono
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stigant wrote:

A similar triangle to the large one is formed by the line from (0,2 2/3) to (2 2/3,2 2/3)

How do you know that the top of the area 2 triangle is (2 2/3, 2 2/3)?

If the base is 6, since the triangle is arbitrary, and the area of the 6 and the 2 is 8, then the height of the combined triangle must follow 8=1/2*6*h, so h = 16/6=2 2/3. Since the height of the 6 is 2, and the width of the 3 is 2, that means the line coming from the origin is at 45 degrees, so if the 6 and 2 combined are 2 2/3 high, the point must also be 2 2/3 over. Hence (2 2/3, 2 2/3)

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stigant
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Ah, I see... Fried Chips is correct in his statement about your calculation.

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Fried Chips
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I'll try and work on a proof, but I'm not sure I'll be able to - the way I found the answer was to assume that all the centre angles were right angles, so I guess the honour of setting the next puzzle goes to the first person who gets the proof.

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bibelot
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File: tri.PNG (3.6 KB)
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Oh, I'll swoop in. See the attached figure. The 6 and 3 triangles can be viewed as having the same height but different bases b and a, so we see that b=2a. Similarly by looking at the 6 and 2 triangles, d=3c. But we can also look at the triangles with areas y and x+3 with bases c and d so that 3y=x+3. Likewise 2x=y+2. We solve to get x=1.8, y=1.6, so the area is 6+3+2+1.8+1.6=14.4.

Fried Chips
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Very slick - I just got a proof, but it's full of complex trigonometry. I won't bother posting it, because a) it's too late, and b) noone would want to read it

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stigant
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bibelot's proof is correct. since fc seems to have conceded, I pass the torch to bibelot.

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bibelot
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Since clearly everybody loves geometry, here's another one.

Let ABC be an isosceles triangle with vertex angle A = 20 degrees. Draw D on side AB such that AD=BC. What is angle DCB?

stigant
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I'll give everybodyelse a chance to answer this one before I reveal the answer.
Edit: whoops, misread the question....


[Edited by stigant at Local Time:03-04-2005 at 01:31 PM]

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DiMono
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stigant wrote:
Ah, I see... Fried Chips is correct in his statement about your calculation.

Yeah, I realised that while watching T.V. tonight. The equation should have been

6:8/3 ~ n+8/3:n
18:8 ~ 3n+8:3n
54n = 24n+64
30n = 64
n=64/30=32/15
h=8/3+32/15=72/15=24/5=4.8
a=1/2*6*4.8=14.4

And that's what happens when you rush.

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Maurog
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DiMono wrote:
Yeah, I realised that while watching T.V. tonight. The equation should have been
6:8/3 ~ n+8/3:n
18:8 ~ 3n+8:3n
54n = 24n+64
30n = 64
n=64/30=32/15
h=8/3+32/15=72/15=24/5=4.8
a=1/2*6*4.8=14.4

And they say TV is bad for for your brain Another urban legend proves to be bogus...

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Fried Chips
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ABC = ACB = 80 degrees

Let length AB be equal to 1:

BC = 2 sin(10) = 0.34

DC^2 = 1^2 + 0.5^2 - 2 x 1 x 0.5 x cos(20) = 0.31
DC = 0.56

sin(80) / 0.56 = sin(DCB) / 0.5
sin(DCB) = 0.88
DCB = 62.1 degrees

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bibelot
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Hmm, I think you made a mistake in your calculations. Plus, the answer should be nicer than that...

Fried Chips
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Ah, misread the question - thought D was the midpoint of AB. New solution:

Let BC = AB = k

k / sin(20) = AC / sin(80)
AC = 2.88k

BD = AC - k = 1.88k

DC^2 = k^2 + (1.88k)^2 - (2 * 1.88)k^2 * cos(80) = 3.80k^2
DC = 1.97k

sin(80) / 1.97k = sin(BDC) / 1.88k
sin(DCB) = 0.94
DCB = 70 degrees

The answer was exact this time, so I hope it's right...

[Edited by Fried Chips at Local Time:03-04-2005 at 06:56 PM]

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