June 2001

The June Six Solutions

Bull's eye 
Charlie will supply the solution here

Two and Two 
Emma will supply the solution here

Muggles Magic 
Chua Zhi Yu, 13 years old, from River Valley High School, Singapore and Andrei Lazanu, age 12, School No. 205, Bucharest, Romania both saw through the deceptiveness of these diagrams and sent explanations of why one square unit of area seems to disappear when the pieces are rearranged.

Look at the triangle with arms of lengths 5 and 13 units. We shall call this triangle T. 
Although the figure at the top looks like this triangle it is not a triangle at all. What appears to be the longest side is not a straight line but actually two lines along the hypotenuses of the red and blue triangles. The gradient or slope of the hypotenuse of the red triangle is 3/8, while the gradient of the hypotenuse of the blue triangle is 2/5. We know 3/8 < 2/5 so the figure at the top has two edges that 'dip inwards' making it into a concave quadrilateral. This shows that the four pieces of this jigsaw fit inside the right angled triangle T leaving a small space uncovered. Exchanging the positions of the red and blue triangles makes these two hypotenuses project outwards enclosing extra area in the shape of a long thin parallelogram.

Adding up the areas, the red triangle has area 12 square units, the blue triangle 5 square units and the other two pieces together 15 square units making a total of 32 square units. 
The right angled triangle T with arms of length 5 and 13 units has area 32.5 square units, an extra half a square unit. By rearranging the pieces the extra area included along the hypotenuse of triangle T is twice this, namely one square unit, and this accounts for the indentation on the bottom of the lower figure.





Gnomon II 
See the file jun01_sixsol505 for the solution here.

Ellipses 
Well done Ryan Catchpole and Prateek Mehrotra, age 14 from Riccarton High School, Christchurch, New Zealand and Andrei Lazanu, age 12, from School 205, Bucharest, Romania for your solutions. 

Ryan noticed that the formula x2 / 36 + y2 / 16 = 1gives a large ellipse crossing through points 6 and -6 on the x axis and points 4 and -4 on the y axis. From this he observed that the formula contains (x2 / 36) and 36 is a square of 6 and -6. When x takes these values then y=0. This also works for the part of the formula 
(y2 / 16) as 16 is a square of 4 and -4. When y takes these values then x=0.

Andrei explained how he found the other equations as follows.
 
I represented first the curve:

							(1)

For this, I observed that both x and y could have values between -1 and +1. I consider x as the independent variable, and from the eq. (1) I determined y:



I gave values to x, from -1 to 1, step 0.1, and I calculated y. I had to calculate two sets of values for y, one corresponding to the plus sign, the other to the minus sign. Then I plotted y as a function of x for both, and I obtained the circle in the middle.

For the curve:

							(2)

I considered again x as the independent variable. It varies between -6 to 6. The equation for y is:



It is visible even from the equation that y varies between -4 and 4.

Now, as I understood that in the general equation:

							(3)

x varies between -a and a, and y between -b and b, I drew all other curves in the same manner. The equations of the other 8 graphs are:

							(4)

							(5)

							(6)

or , i.e. a circle of radius 4.

							(7)

							(8)

							(9)

							(10)

							(11)


All these ellipses are symmetrical about both x and y axes because by changing x to -x and/or y to -y the equation (3) doesn't change.

Click here for the programs Andrei wrote in Matlab for Windows to represent the ellipses:

Note for whoever marks this one up : Please put these programs on a separate page with a hyperlink to 'Click here'

%circles and ellipses
%circle of radius 1
x=[-1:0.05:1];
y=sqrt(1-x.*x);
y1=-sqrt(1-x.*x);
plot(x,y,'b',x,y1,'b')
hold
clear

%ellipse x^2/36+y^2/16=1
x=[-6:0.1:6];
y1=sqrt(16*(1-x.*x/36));
y2=-sqrt(16*(1-x.*x/36));
plot(x,y1,'r',x,y2,'r')
grid
clear

%ellipse x^2/7^2+y^2/16=1
x=[-7:0.1:7];
y1=sqrt(16*(1-x.*x/7^2));
y2=-sqrt(16*(1-x.*x/7^2));
plot(x,y1,'c',x,y2,'c')
clear

%ellipse x^2/5^2+y^2/16=1
x=[-5:0.1:5];
y1=sqrt(16*(1-x.*x/5^2));
y2=-sqrt(16*(1-x.*x/5^2));
plot(x,y1,'m',x,y2,'m')
clear

%ellipse x^2/4^2+y^2/16=1
x=[-4:0.1:4];
y1=sqrt(16*(1-x.*x/4^2));
y2=-sqrt(16*(1-x.*x/4^2));
plot(x,y1,'y',x,y2,'y')
clear

%ellipse x^2/3^2+y^2/16=1
x=[-3:0.1:3];
y1=sqrt(16*(1-x.*x/3^2));
y2=-sqrt(16*(1-x.*x/3^2));
plot(x,y1,'g',x,y2,'g')
clear

%ellipse x^2/2^2+y^2/16=1
x=[-2:0.1:2];
y1=sqrt(16*(1-x.*x/2^2));
y2=-sqrt(16*(1-x.*x/2^2));
plot(x,y1,'k',x,y2,'k')
clear

%ellipse x^2/1^2+y^2/16=1
x=[-1:0.1:1];
y1=sqrt(16*(1-x.*x/1^2));
y2=-sqrt(16*(1-x.*x/1^2));
plot(x,y1,'--b',x,y2,'--b')
clear

%ellipse x^2/7^2+y^2/1=1
x=[-7:0.1:7];
y1=sqrt(1*(1-x.*x/7^2));
y2=-sqrt(1*(1-x.*x/7^2));
plot(x,y1,'--r',x,y2,'--r')
clear

%ellipse x^2/6^2+y^2/9=1
x=[-6:0.1:6];
y1=sqrt(3^2*(1-x.*x/6^2));
y2=-sqrt(3^2*(1-x.*x/6^2));
plot(x,y1,'--c',x,y2,'--c')


Parabella 
Charlie will supply the solution here