| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 896.1 | | HPSTEK::XIA | | Thu Jul 21 1988 18:07 | 6 |
| I agree that this problem is hard as it is. However, if we turn
it into a problem in analytic geometry, then it becomes a matter
of messy algebraic manipulation, but it will give the correct answer
in the end.
Eugene
P.S. Is it considered cheating for using analytic geometry?
|
| 896.2 | | ZFC::DERAMO | Hello, world\n | Thu Jul 21 1988 20:05 | 48 |
| Re .1:
>> P.S. Is it considered cheating for using analytic geometry?
Why should it be considered cheating? Isn't that how
everyone solves these? :-)
The title is "A Geometry Problem" but I don't see that
as a "requirement" on the proof.
Dan
P.S.
The following claim was made without proof. Should you
believe it? Let your own judgement be your guide. :-)
From: BRIANE::ASHBY::USENET "USENET Newsgroup Distributor 14-Jul-1988 1820" 14-JUL-1988 18:26:06.21
To: @SUBSCRIBERS.DIS
CC:
Subj: USENET sci.math newsgroup articles
Newsgroups: sci.math,rec.puzzles
Path: decwrl!purdue!bu-cs!bloom-beacon!tut.cis.ohio-state.edu!osu-cis!att!chinet!mcdchg!clyde!watmath!watdragon!dmnhieu
Subject: Re: A Geometry Problem
Posted: 8 Jul 88 15:59:58 GMT
Organization: U. of Waterloo, Ontario
Xref: decwrl sci.math:4433 rec.puzzles:1697
In article <[email protected]> [email protected] (Krishna) writes:
>I came across the following geometry problem and found it quite challenging.
>
>Take any convex quadrilateral ABCD. Construct four isosceles right triangles
>using the sides as the hypotenuses. The legs are outside the the
>quadrilateral. Call the new point M, N, O, and P in some sort of clockwise
>order.
>
>Prove: MO = NP and MO is perpendicular to NP. Does ABCD have to be convex?
>
>Ambati Jayakrishna
>[email protected]
>[email protected]
ABCD does not have to be convex, (by using the new Maple's geometry package),
also the triangles can be all inside the quadrilateral ABCD.
[email protected]
|
| 896.3 | Treat as Vectors, Get Lengths and Dot Product | ATLAST::FRAZER | Je suis prest! | Fri Jul 22 1988 00:47 | 96 |
| Think of A,B,C and D as vectors from (0,0)
A.
.P / \ .M
/ \
/ \
D. .B
\ |
\ |
\ |
. . .N
\ |
\ |
.O \|
.C
Quadrilateral ABCD
Vectors A=(Xa,Ya)
Vectors B=(Xb,Yb)
Vectors C=(Xc,Yc)
Vectors D=(Xd,Yd)
CCW(Xv,Yv) = (-Yv,Xv) ! rotate a vector counter clocwise
AB = B - A = (Xb-Xa,Yb-Ya) !Vector from point A to B
M = A + AB/2 + ccw(AB/2)
M = (2Xa + Xb -Xa +Ya -Yb, 2Ya +Yb - Ya +Xb -Xa)
--------------------------------------------
2
M = (Xa +Xb +Ya -Yb, Ya +Yb +Xb -Xa)
--------------------------------
2
likewise:
N = (Xb +Xc +Yb -Yc, Yb +Yc +Xc -Xb)
--------------------------------
2
O = (Xc +Xd +Yc -Yd, Yc +Yd +Xd -Xc)
--------------------------------
2
P = (Xd +Xa +Yd -Ya, Yd +Ya +Xa -Xd)
--------------------------------
2
MO = O-M = (Xc +Xd +Yc -Yd -Xa -Xb -Ya +Yb, Yc +Yd +Xd -Xc -Ya -Yb -Xb +Xa)
-----------------------------------------------------------------
2
NP = P-N = (Xd +Xa +Yd -Ya -Xb -Xc -Yb +Yc, Yd +Ya +Xa -Xd -Yb -Yc -Xc +Xb)
-----------------------------------------------------------------
2
4|MO|^2 = (Xc +Xd +Yc -Yd -Xa -Xb -Ya +Yb)^2 +
(Yc +Yd +Xd -Xc -Ya -Yb -Xb +Xa)^2
=
+XaYd -XaXc -XaYb +XaXa
-XbXd -XbYc +XbXb +XbYa
+XcXc -XcYd -XcXa +XcYb
+XdXd +XdYc -XdXb -XdYa
-YaXd -YaYc +YaXb +YaYa
+YbXc -YbYd -YbXa +YbYb
+YcXd +YcYc -YcXb -YcYa
+YdYd -YdXc -YdYb +YdXa
similarly
4|NP|^2=
+XaXa -2XaXc -2XaYb +2XaYd
+XbXb -2XdXb +2XbYa +2XcYb -2XbYc
+XcXc +2XdYc -2XcYd
+XdXd -2XdYa
+YaYa -2YcYa
+YbYb -2YdYb
+YcYc
+YdYd
The two sum of squares are equal, thus the lengths are equal.
Now the dot product
MO . NP :
(Xc +Xd +Yc -Yd -Xa -Xb -Ya +Yb, Yc +Yd +Xd -Xc -Ya -Yb -Xb +Xa) .
(Xd +Xa +Yd -Ya -Xb -Xc -Yb +Yc, Yd +Ya +Xa -Xd -Yb -Yc -Xc +Xb)
=
+(Xc +Xd +Yc -Yd -Xa -Xb -Ya +Yb)(Xd +Xa +Yd -Ya -Xb -Xc -Yb +Yc)
+(Yc +Yd +Xd -Xc -Ya -Yb -Xb +Xa)(Yd +Ya +Xa -Xd -Yb -Yc -Xc +Xb)
which, if you expand it winds up all cancelling out to 0.
Thus the vectors are perpendicular.
QED.
|
| 896.4 | | HPSTEK::XIA | | Fri Jul 22 1988 11:34 | 15 |
| re .3
Congratulations for the solutions. There is a possible simplification
though. You can assume one of the sides of the quadrulateral to
be on X-axis or Y-axis. This might simplify the solution process
(maybe).
re .2
When I was in high school, plane geometry was a favorate subject
amond the students in my school since it requires
luck/insight/persisitence. Later when we learnt analytic geometry,
trig and vector analysis, these techniques took the fun out of
the geometry problems (actually not really, but you know what I
mean). So we sort of refuse to use these techniques in the "fun"
problems. Oh well.
Eugene
|
| 896.5 | | ZFC::DERAMO | Hello, world\n | Fri Jul 22 1988 21:24 | 7 |
| Yes, .3 was well done. Re .4, I do understand what you
mean. If there is "extra credit" for a geometric solution,
then there is a desire to do it that way. On the other
hand, if points are taken off for doing more work than
is necessary ....
Dan
|
| 896.6 | | CHALK::HALLYB | The smart money was on Goliath | Sun Jul 24 1988 14:46 | 12 |
| One of my math professors, Henry Mann, had an interesting attitude
about what you use for proofs. If you have an easy proof that uses
the Axiom of Choice versus a difficult proof that is independent
of the Axiom of Choice, then you go for the more difficult one.
Because then "You have proved more", for your proof holds in a wider
space of mathematical systems.
It would seem, then, that a purely geometric proof would be superior
by that, uhhh, metric. At least I *think* analytic geometry requires
more foundation than plane geometry.
John
|