Math/Geometry puzzle from china
mr_gadget
Today I found and solved an interesting math/geometry problem which is
supposedly high school level, but gave me a hard time. I wanted to post it
here for someone to solve so I can compare and find out if my answers are
right!
And I found it on BBC of all places! Search revealed no other discussions of
it. I am afraid its based on a picture so you will have to go the article on
BBC and scroll to the end to see it.
http://news.bbc.co.uk/2/hi/uk_news/education/6589301.stm
Yes I am talking about the chinese prism one, not the other stupid british
one :-|
Mr gadget
Brian Tung
> Today I found and solved an interesting math/geometry problem which is
> supposedly high school level, but gave me a hard time. I wanted to post it
> here for someone to solve so I can compare and find out if my answers are
> right!
>
> And I found it on BBC of all places! Search revealed no other discussions of
> it. I am afraid its based on a picture so you will have to go the article on
> BBC and scroll to the end to see it.
>
> http://news.bbc.co.uk/2/hi/uk_news/education/6589301.stm
I'm puzzled on (i) where they want you to prove that BD is perpendicular
to A1C. As far as I can tell, those two lines are skew. So they can't
be perpendicular. Projected onto ABCD, they're perpendicular, but
projected onto ADD1A1, they meet at an angle of arctan(4sqrt(3)/3).
A similar comment goes for (iii) which states "Determine the angle
formed by lines Ad and BC1, which are in different planes." Well, if
they're in different planes, how is the angle between them defined? If
you project them into the plane of ABCD, then they are perpendicular.
If you project them into the plane of ADD1A1, then they're parallel. So
I really don't understand what they're asking for.
Also, what does "foot of perpendicular" mean? Is it that E is the
intersection of AC and BD?
At any rate, from AD = 2, DC = 2sqrt(3) and AD perp DC, we see that ADC
is a 30-60-90 triangle, so AC = 4. And since AC perp BD (intersecting
at E, presumably), we see that AED and DEC are also both 30-60-90
triangles, with AE = 1 and EC = 3. Then since AB = 2, AEB must also be
a 30-60-90 triangle, with EB = ED = sqrt(3), and BC = DC = 2sqrt(3).
Thus, quadrilateral ABCD is a kite.
Since AA1 = sqrt(3), we have A1AE a 30-60-90 triangle, so AE = 2. With
EC = 3 and CC1 = AA1 = sqrt(3) (presumed?), ECC1 is also a 30-60-90
triangle, with EC1 = 2sqrt(3), and since A1C1 = AC = 4 (again presumed),
A1C1E is yet another 30-60-90 triangle, so planes A1BD and BC1D meet at
right angles. I'm sure that didn't come as any surprise, given the
preponderance of 30-60-90 triangles. This answers (ii).
Maybe you can clarify what they're asking for in (i) and (iii). I think
I must be missing something (maybe if I had the Chinese original?), but
I can't think what.
--
Brian Tung <br...@isi.edu>
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Mark Brader
> As far as I can tell, those two lines are skew. So they can't
You can define an angle between two skew lines easily enough: each
one has a unique point where it comes nearest to the other, so just
view them along the line connecting those two points. In other words,
draw a parallel to one line through the point of closest approach on
the other line, and measure the angle using that new line.
I assume that was the intent here. I haven't seen it done before either.
--
Mark Brader | "[Your orders are] to figure out what I would have ordered
m...@vex.net | you to do, if I really understood the situation ... [and]
Toronto | to follow those orders I hypothetically would have given."
-- Shan (John Barnes, "Earth Made of Glass")
My text in this article is in the public domain.
Simon Tatham
> I'm puzzled on (i) where they want you to prove that BD is perpendicular
> to A1C. As far as I can tell, those two lines are skew. So they can't
> be perpendicular. Projected onto ABCD, they're perpendicular, but
> projected onto ADD1A1, they meet at an angle of arctan(4sqrt(3)/3).
>
> A similar comment goes for (iii) which states "Determine the angle
> formed by lines Ad and BC1, which are in different planes." Well, if
> they're in different planes, how is the angle between them defined?
Same answer in both cases. If you insist that an angle can only be
measured where two lines or planes meet, then don't project:
_translate_ one until it meets the other, and then measure the angle
between them. Under this definition, BD and A1C are perpendicular.
(This definition arises naturally if you approach the entire problem
via coordinate geometry: take the inverse cosine of the dot product
of two unit vectors, and you'll find the angle between them,
regardless of whether the lines along which the vectors pointed were
originally touching or not.)
--
Simon Tatham "_shin_, n. An ingenious device for
<ana...@pobox.com> finding tables and chairs in the dark."
Brian Tung
> Same answer in both cases. If you insist that an angle can only be
> measured where two lines or planes meet, then don't project:
> _translate_ one until it meets the other, and then measure the angle
> between them. Under this definition, BD and A1C are perpendicular.
Of course. Thanks for clarifying.