japanese entrance exam problem
TRANSCRIPT
Japanese entrance exam problemAuthor(s): Gyula DarvasiSource: The Mathematics Teacher, Vol. 94, No. 2 (February 2001), p. 105Published by: National Council of Teachers of MathematicsStable URL: http://www.jstor.org/stable/20870595 .
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(Continued from page 83)
?o the roots are to be found in the interval (-100; 100), which is somewhat unwieldy. The theorem gives
M f 2 J? ,fn{ M = max<??J?? 3?,4?
[0.2 V0.2 V0.2 V0.2J = max{l0,
, 2154,2.723}
= 10.
The roots therefore lie in (-20, 20), a considerable improvement.
Jens Carstensen DK-2000 Frederiksberg Denmark
Japanese entrance exam problem The original problem in "Japanese High School Entrance Examina tions" on pages 30
- 35 in the January 2000 issue of the Mathe matics Teacher reads as follows:
In the parallelogram ABCD, E and Fare the midpoints of CD and AD, respectively. G is the point of intersection of AE and BF. Express the ratio of the areas of triangle EFG and triangle BCE.
See figure 1 (Darvasi).
a_F_
? c
Fig. 1 (Darvasi) AEFG:ABCE = 3:10
Taking one unit for the area of parallelogram ABCD and draw ing diagonals AC and BD, we obtain the results that area AFDE = 1/8 and area ABCE = 1/4 Denoting by x, u, v, and w the areas of triangles EFG, BEG, ABG,m?AFG, we get the follow ing equations: ?rea ABCE = area AABF
= areaAAD? = ?
area ABEF = x + u=*
3
area AADE = x^
+ w, ?
area AABF = + w J
=>i? = x + |,
and
area Ai4J5F = x + ?; = i
By drawing perpendicular seg ments from F to AE and from jB to AE, respectively, we can con clude that
x=EG~? w AG"
=$x*v = u*w,
which enables us to make the fol lowing computation:
?\x + - = ?X ? ?X
2 1 3 1 3 * X + -X = ?-X?x+x
8 64 8 8
5 = _3_ 8X 64
3 x = ?
40
Therefore, the desired ratio is
-:i = 3:10. 40 4
Gyula Darvasi darvasi@agy. bgytf.hu Ny?regyh?za College Sostoi ut 311b Hungary
A dissection proof In "A Geometry Solution from
Multiple Perspectives," on pages 324-27 of the April 2000 issue of the Mathematics Teacher, Phillip Nissen offers four different solu tions to the following problem: WXYZ is a square, with M the midpoint of WZ; the lines XZ and YM partition the square into four portions marked P, Q,
R, and S. Find... the ratios of the areas? :Q:E:S. See fig ure 1 (Hoehn).
X Y
W
Fig. 1 (Hoehn)
I have another synthetic proof that differs from the one presented in the article. This one might be called a dissection proof. Let Lt 0, and JV be the respective mid points o?WX, XT, and YZ, and construct the indicated segments shown in figure 2 (Hoehn).
Fig. 2 (Hoehn) Construct midpoints L, , N.
Since WN and YM are medians of AWZY and since the medians of a triangle are concurrent, ZK is also a median. Since the three
medians of any triangle divide the triangle into six smaller tri angles of equal area, each smaller triangle in AWZY has an area of P. The same is true for AWXY. Therefore,P:?:?:S = P:2P: 4P:5P=1:2:4:5.
Larry Hoehn [email protected] Austin Peay State University Clarksvitte, TN 37044
Another geometr?e perspective I was just trying my hand at the problem offered in "A Geometry Solution from Multiple Perspec tives" (Mathematics Teacher 93 [April 2000]: 324-27). After find ing the solution, I continued reading the article, only to find that I had discovered a different geometric solution. Like the syn thetic approach, my solution is based on similar triangles, but it obviates the need for an interme diate step.
I first broke the original square into sixteen congruent
squares and recognized that one of those smaller squares was similar to that in the original problem. See figure 1 (Mutford).
R'
Fig. 1 (Muford) Similar triangle approach
The small box in row 3, column 3 has regions P' = (1/16)P,Q' =
(1/16)Q,?' = (1/16)?, andS' =
(l/16)S.Thus,P:Q:?:S =
FiQ'-.R'-.S". Let us use these
boxes to represent regions P, Q, R, and S; so doing is nothing more than counting boxes, half boxes, or quarter-boxes. Then
= (3/4X1/16) + (2/4X1/16) + F
= (5/64) + ', P-P" = 5/64
16P'-P' = 5/64; F = 5/960
of the total area.
Q = (1/16) + (3/4X1/16) + (1/4X1/16) + (2/4X1/16) + Q'
Q = (10/64)+ Q', 16e'-Q' = 10/64;
Q' = 10/960.
R = 3(1/16) + 2(2/4X1/16) + (1/4X1/16) +(3/4X1/16)+ ?'
? = (20/64) + R', m'-R' = 20/64;
?' = 20/960.
S = 5(1/16) + 2(2/4X1/16) + (1/4X1/16) + S'
S =(25/64) +S', 16S'-S' = 25/64;
S' = 25/960.
Thus,
F:Q':R':S' = P:Q:R:S = 5:10:20:25 = 1:2:4:5.
I plan to put this problem on my challenge board to see what my students come up with.
Jason Mutford [email protected]
Margaretville Central Schools Margaretville, NY 12455
(Continued on page 111)
Vol. 94, No. 2 ? February 2001 m
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