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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

106.0. "Geometry Problem IV" by HARE::STAN () Mon Jul 30 1984 03:13

Newsgroups: net.math,net.puzzle
Path: decwrl!decvax!tektronix!uw-beaver!cornell!vax135!houxz!houxm!mhuxl!ulysses!unc!mcnc!ecsvax!pizer
Subject: O.K. Here's another problem
Posted: Thu Jul 12 15:23:41 1984

I wasn't going to post a problem tonite, but since I haven't solved this one
yet, it will be more of challenge for myself, sort of a race against time.
Since it is a multiple choice question, the answers are listed below.  DON'T
GUESS!  If you do come up with an answer, and then mail it to me, make sure
you label it Geometry IV solution, so I don't cheat accidently.  Here it is:
Triangle ABC (scalene) has area 10.  Points D, E, and F, all distinct from
A, B and C, are on sides AB,BC and CA respectively, and AD=2, DB=3.  If
triangle ABE and quadrilateral DBEF have equal areas, then that area is

(A) 4   (B) 5   (C) 6   (D) 5/3 sqrt(10)   (E) not uniquely determined

HINT:  National Math Exam questions rarely end up being not uniquely
determined (that doesn't mean it isn't, just that they rarely are)
Good luck!

Billy Pizer
North Carolina School of Science and Mathematics
({decvax,akgua,etc...}!mcnc!ecsvax!pizer)
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106.1HARE::STANMon Jul 30 1984 03:1350
Newsgroups: net.math
Path: decwrl!dec-rhea!dec-tonto!luong
Subject: Geometry IV puzzle solution.
Posted: Mon Jul 16 13:35:35 1984

Problem:
-------
>Consider points D,E,F on sides AB,BC,and CA of scalene triangle ABC, with
>D,E,F, all distinct from A,B,and C. If area of ABC = 10, AD = 2, DB = 3, and
>area of ABE = area of DBEF, what is area of ABE?

Terminology:	Let's denote the area of a figure by using brackets, eg:
-----------		(PQR) =  area of triangle PQR

Solution:		(AEC) = (ABC) - (ABE) 
--------
Since (ABE)=(DBEF):	(AEC) = (ABC) - (DBEF)       

Therefore:		(AEC) = (EFC) + (DAF)        (1)

But we also have:	(AEC) = (EFC)+ (EAF)	     (2)

Comparing (1) and (2) gives:     (DAF) = (EAF)       (3)

Since DAF and EAF share the common base AF, (3) shows that D and E are
at the same distance from AF, ie. that lines DE and AC are parallel.

As DE and AC are parallel, BE = BD  = 3
			  ---   ---  ---
			   BC   BA    5

But (ABE) = BE   therefore  (ABE) = (ABC) * BE  = 10 * 3  = 6  .
    ----   --- 		       		   ---        ---
    (ABC)   BC				    BC	       5


NB. Our colleague who posted this puzzle stated that it is from the National
Math Exam. Since I have been in the USA only since January 1984, could someone
please explain to me what is the National Math Exam? How many years of school
do you have to complete before taking this exam?

In the school system in France, this would be a medium difficulty problem
for Year 9 (ie. 3 years before secondary school graduation). A kid has to do
a total of 12 years (5 years in primary school, and 7 in secondary school)
before going to college.

Van Luong Nguyen,
Digital Equipment Corporation,
Nashua, NH.