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Topic: Some help with problems
Replies: 4   Last Post: Jun 22, 2013 3:52 AM

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Posts: 11,749
Registered: 12/4/04
Re: Some help with problems
Posted: May 29, 2013 5:09 PM
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On Wednesday, May 29, 2013 1:19:33 PM UTC-5, stony wrote:
> Hi,
> I have a set of 20 problems that my daughter has to solve as practice.
> We are done with 17 of them, but the following has us stumped with
> trying to apply the right rules. Any help would be appreciated.
> 7. A parallelogram has integer side-lengths and diagonals 40 and 42.
> What is the area. They call it a parallelogram, but I say the
> diagonals have to be the same. I say its a rhombus, but the teacher
> says parallelogram. So, now I cannot make assumptions about the
> angles and other assumptions based on that. Any idea how to solve
> this? Rhombus is a type of parallelogram? Any clues or ideas are
> appreciated.

Parallelogram means that opposite sides are parallel; so a rhombus is a type of parallelogram.

But so is something like this:

/ /

and you can see that it is not a rhombus, and the two diagonals are not the same.

The problem specifically says that the two diagonals are NOT the same, so I do not understand what it is you mean by "diagonals have to be the same."

Are you familiar with the "parallelogram law"? It relates the lengths of the sides to the lengths of the diagonals. If the lengths of the sides are a and b, it should allow you to find the value of a^2 + b^2, and from there using the fact that a and b are integers will likely determine their values.

> 5. How do we prove (3/2) > ( (2^571 +3) / (3^247 + 2)). Numerical
> calculations are too big and out of range.

Cross multiplying you have that the inequality holds if and only if 3(3^{247}+2) > 2(2^{571}+3), which becomes 3^{248}+6 > 2^{572}+6, which in turn is equivalent to 3^{248} > 2^{572}. So it comes down to testing this inequality.

You can now use logarithms, or back-of-the-envelope computations; however, my computations seem to suggest the inequality is NOT true...

Arturo Magidin

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