It seemed as if all of South Fremantle Senior High School in Australia put up Web pages illustrating solutions to the problem this week, and it was hard to choose among them! Following are three of the earliest and best submissions, featuring both discussions and drawings. Indiana, Hawaii, and Dartmouth '92 are also represented.
Chid Gilovitz
Grade: 8
School: South Fremantle Senior High
Hi! I wrote a web page on the July 22-26 POW.
You will find it at gilovitc7.22.96.html.
Seeya!
Chid
Heather-Jayne Kirkby
Grade: Grade 8
School: South Fremantle Senior High School
This is the first time I have entered any of your problems
although I have figured out and completed many. I find most
of them very challenging and interesting. I hope to submit
more of my answers to your problems in the future.
You will find my entry at kirkby7.22.96.html.
Thank you for supplying me with the problem. See you next time!
Heather-Jayne
Paul Malone
Grade: 9
School: South Fremantle Senior High School
I have never before sent an entry in to these problems so
I am submitting this solution. In the past I have received
a printed copy of many problems but I never completed a
solution within the time limit. I have them all presented
in a file, some completed and some incomplete, so I can look
back when I am stuck. This week I finished because my
mathematics teacher gave us time in class. My solution is
not terribly accurate but I think that it will suffice. You
will find it at malone7.22.96.html.
I hope you find the answer and my working out all right.
Bryce Johnson
Grade: between 10 and 11
School: Center Grove High School, Greenwood, Indiana
Because the side of an equilateral triangle is longer than the
altitude, the square adjacent to the side is the larger square.
Therefore the side of the equilateral triangle is radical 56,
or 2 radical 14.
The altitude creates a 30-60-90 right triangle with the altitude
as the longer leg. Dividing by 2 and multiplying by radical 3
gives the length of the altitude as the square root of 42, which
means the area of the smaller square is 42.
Jason Yeung
Grade: 9 / 10
School: Iolani School
The area of the smaller of these squares is 42.
First of all, in an equilateral triangle, if the side is 2x,
then the altitude of the triangle is x* sqrt 3. We can prove
that because an equilateral triangle is composed of two
30-60-90 right triangles.
Since 2x is larger than x*sqrt 3, the area of the square with the
side of the triangle as its side is bigger than the one with the
altitude.
If the area of that square is 56, then its side will be sqrt of 56,
or 2* sqrt of 14, which is also the length of the side of the
triangle.
If 2x (the side) is 2* sqrt of 14, then x is sqrt of 14, and
x*sqrt of 3 is sqrt of 42, which is the altitude and the side of
the smaller square.
Since the side of the smaller square is sqrt 42, the area of the
square is sqrt of 42 * sqrt of 42, which is 42, the answer to the
question.
Brian Gordon
Dartmouth '92
The altitude of an equilateral triangle divides the triangle
into 30-60-90 triangles. Therefore, it's easy to see that
the ratio between the sides of the squares is sqrt(3) : 2.
Squares are all similar, and the ratio of areas of similar
figures is equal to the square of the ratio of corresponding
sides. Squaring sqrt(3) : 2 gives an area ratio of 3:4
Since the big square has ratio 56, we have 3:4 = x:56
So the area of the little square is 42.
--bri