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ppauper
Posts: 70239
Joined: November 15th, 2001, 1:29 pm

### Re: I can't solve this ...

area = $\frac{1}{2}$ base $\times$ height

it's one of those things you learn when you are what, 6?
the upright splits the big triangle into 2 right triangles with bases say b_1 and b_2 with b_1+b_2=b
each of those right triangles has an area equal to 1/2 of the corresponding rectangle with sides b_1 and h for one triangle and b_2 and h for the other
so area =1/2 b_1 h +1/2 b_2 h =1/2 (b_1+b_2) h =1/2 bh
..you're changing the problem again! This is a different problem, one that doen't satisfy one of the given constraints!
no it isn't, the base times height formula has no requirement that the sides have real rather than complex lengths

outrun
Posts: 4573
Joined: April 29th, 2016, 1:40 pm

### Re: I can't solve this ...

H = 6?

The height of this gray rectangle is larger than the 6, right? If you rotate the purple line to the left around it's lower left origing then the top of the purple line will start to show a gap below the white line that moves to the top left corner. Hence the area of the rectangle > 60, hence the triangle > 30,

Any good teacher would point out this sanity check

Posts: 23951
Joined: September 20th, 2002, 8:30 pm

### Re: I can't solve this ...

H = 6?

The height of this gray rectangle is larger than the 6, right? If you rotate the purple line to the left around it's lower left origing then the top of the purple line will start to show a gap below the white line that moves to the top left corner. Hence the area of the rectangle > 60, hence the triangle > 30,

Any good teacher would point out this sanity check
But the width of the gray rectangle is SMALLER than the 10, right? If you rotate the white line to the top around it's upper left origin then the right edge of the white line will start to extend past the rectangle and the top right left corner. Hence the area of the rectangle < 60, hence the triangle < 30,

So how do the two line-to-side ratios differ and does the effect of the purple line outweigh the effect of the while line?

The answer is that they exactly balance each other. The height of the rectangle is 6/cos(theta) and the width of the rectangle is 10*cos(theta) where theta is the narrow angle found in both the large and small triangles.

Thus the rectangle is exactly 6*10 and the triangle is exactly 30.

outrun
Posts: 4573
Joined: April 29th, 2016, 1:40 pm

### Re: I can't solve this ...

Yes, the width is smaller! I was just going to retract it! Back to the drawing board

Posts: 23951
Joined: September 20th, 2002, 8:30 pm

### Re: I can't solve this ...

Yes, the width is smaller! I was just going to retract it! Back to the drawing board
We'll wean you of your Russian tendencies yet!

outrun
Posts: 4573
Joined: April 29th, 2016, 1:40 pm

### Re: I can't solve this ...

It's like
A=3, B=10, A+B=20, what is A*B? So dropping different constraints will give different outcomes.

It's not easy to do this while carrying an Ikea desk while biking, so .. I'll be back later!

outrun
Posts: 4573
Joined: April 29th, 2016, 1:40 pm

### Re: I can't solve this ...

What's this?? I'm not allowed to have a sad face while standing on a pile of Ikea planks?!
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ppauper
Posts: 70239
Joined: November 15th, 2001, 1:29 pm

### Re: I can't solve this ...

the top one, my guess would be you need 2 people
the bottom one is cropped so I can't see it all

Posts: 23951
Joined: September 20th, 2002, 8:30 pm

### Re: I can't solve this ...

What's this?? I'm not allowed to have a sad face while standing on a pile of Ikea planks?!
You need a clone and a pencil.

Paul
Posts: 10093
Joined: July 20th, 2001, 3:28 pm

### Re: I can't solve this ...

Clearly maths education in Russia is so awful that children are taught to check every question for consistency/errors. Or maybe Russians just aren't very trusting.

Whatever way you look at this Americans come out as far superior to Russians.

Posts: 23951
Joined: September 20th, 2002, 8:30 pm

### Re: I can't solve this ...

The Russian interpretation is a reaction to decades of Communist Party rule -- if the authorities say "X", then "not(X)" must be true.

outrun
Posts: 4573
Joined: April 29th, 2016, 1:40 pm

### Re: I can't solve this ...

Here's proof that area must be at least 36, with a height of 6 the base must be at least 12.

If you keep this up I'm going to report you to trump, tell him that your arguing with his russian friends.
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Cuchulainn
Posts: 61185
Joined: July 16th, 2004, 7:38 am
Location: Amsterdam
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### Re: I can't solve this ...

yes, h=6
yes, h = 6.
http://www.datasimfinancial.com
http://www.datasim.nl

Every Time We Teach a Child Something, We Keep Him from Inventing It Himself
Jean Piaget

Posts: 23951
Joined: September 20th, 2002, 8:30 pm

### Re: I can't solve this ...

6A + 6/A = 10
6A^2 -10A + 6 = 0
A = 0.833333 + 0.552771i

So the lengths of the two halves of the 10-unit hypotenuse created by the 6-unit perpendicular would be 5 + 3.31662i and 5 - 3.31662i.

outrun
Posts: 4573
Joined: April 29th, 2016, 1:40 pm

### Re: I can't solve this ...

So the area is *always* 30 but the figure is impossible, it can't be drawn because the two triangles both have two complex length sides and one real side of 6, this makes e.g. the height and width of the grey area in the OP also complex, etc etc. However if you allow for imaginary lengths and areas then the complex terms of the imaginary areas of the two sub-triangles always cancel out. This is unexpected for me.

If you require lengths and areas to be real values then you can show that it's an impossible figure, like I did by showing that a subset of the stated properties results is the length 10 hypo to have a minimal length of 12, hence an internal inconsistency.

How do you draw a line segment with a complex length? Can quaternions coordinates have complex distances between them?