CHECK: Suppose there were 79 coconuts in the original pile.
• The first man divided the 79 coconuts into 3 equal groups of 26 coconuts , with
one left
over for the monkey. He took 26 coconuts and left the other 52 in the pile.
• The second man divided the 52 coconuts into 3 equal groups of 17 coconuts,
with one
left over for the monkey. He took 17 coconuts and left the other 34 in the pile.
• The third man divided the 34 coconuts into 3 equal groups of 11 coconuts, with
one left
over for the monkey. He took 11 coconuts and left the other 22 in the pile.
• In the morning, the men divided the 22 coconuts into 3 equal groups of 7
coconuts, with
one left over for the monkey.
So, it works! Finally, we count up the number of coconuts each man received:
• The first man ends up with x1 + x4 = 26 + 7 = 33 coconuts.
• The second man ends up with x2 + x4 = 17 + 7 = 24 coconuts.
• The third man ends up with x3 + x4 = 11 + 7 = 18 coconuts.
COMMENT: If one knows modular arithmetic , some of these calculations can be
simplified
(or skipped altogether). For example - we could rewrite the equation relating N
to x4 as
follows: 8N - 65 = 81x4. This indicates that 8N -65 is a multiple of 81, or - in
the language
of modular arithmetic - 8N ≡ 65 (mod 81). (This is read aloud as \8N is
congruent to 65
(mod 81)". By properties of modular arithmetic, it is relatively easy to reduce
this congruence
to: N ≡ 79 (mod 81). This means that N is a solution to the original problem if
(and only
if) N -79 is a multiple of 81. That means 79, 160, 241, 322, etc. would satisfy
the conditions
of the problem (aside from the \less than 100 coconuts" condition, which is what
results in a
unique solution).
4. In the figure shown below, point A is at the origin, point B is at (4,0), and
point C is at (3,2).
The triangles ∆ ACE, ∆ CBD and ∆ ABF are all equilateral.
(a) Find the sum of the distances from point (2,1) to each of the vertices of
∆
ABC.
(b) Lines
all intersect at a single point. Find the coordinates of this point .
(c) Find the sum of the distances from this intersection point to each of the
vertices of
∆ ABC.
(a) Using the distance formula, we have:
• Distance to (0,0):
• Distance to (3,2):
• Distance to (4,0):
So, the sum of these distances is
(or approximately 5.886).
(b) We're given that all three lines meet at a single point. This is true
(though by no means
obvious), which means that we need only find the intersection point of any two of
these
lines. To do this, we'll need to find the equations of two of the lines. Here,
we will find
the equations of
and
,
by finding the coordinates of points F and D and then
applying the point-slope equation for a line.
• Finding F and
Since ∆ ABF is an equilateral triangle with side length 4, it is not difficult
to find
the coordinates of F - we need only draw an altitude from F to the midpoint of
This creates a 30-60-90 right triangle with hypotenuse length 4, thus, the
altitude's
length is
and so the coordinates of F are
It fol lows that the slope of
is
Therefore, we have the following equation for
• Finding D and
First, notice that
passes through the point (0,0). Therefore, once we find the
coordinates (a, b) of point D, it will immediately follow that the equation of
line
is just
The distance formula gives us
therefore,
∆ BCD is an equilateral triangle
with side length
Therefore, D(a, b) is a point that satisfies both of the following
distance equations:
(a - 3)2 + (b - 2)2 = 5 (since D is distance
from (3,2), and
(a - 4)2 + (b - 0)2 = 5 (since D is distance
from (4,0)
This is a system of two equations in two variables which can be solved
algebraically.
(Details omitted { the algebra is straightforward, but takes several lines to
complete.)
Your result should be:
Therefore, the equation of line
is
This can be simplified somewhat by rationalizing the denominator, we may multiply
the numerator and denominator each by
to get the equation
for
Now, we can find the intersection point of the lines by solving the system of
equations:
After quite a bit of algebra, we find that the solution to this system is
Or, as decimal approximations , x ≈ 2.818 and y ≈ 1.005.
(c) We will now find the distance from this intersection point to each of the
vertices of the
triangle.
• Point A:
• Point B:
• Point C:
So, the sum of these distances is approximately 5.55.
