### Video Transcript

The diagram shows a uniform lamina
π΄π΅πΆ from which a triangle πΊπ΅πΆ has been cut out. π΄π΅πΆ was an equilateral triangle
with a side length of 93 centimeters and center of mass πΊ. Find the coordinates of the new
center of mass of the resulting lamina. Round your answer to two decimal
places if necessary.

The statement refers to this
arrow-shaped lamina formed from the equilateral triangle π΄π΅πΆ with the triangle
πΊπ΅πΆ cut out of it. The specific information weβre
given is that π΄π΅πΆ has a uniform mass distribution, its side length is 93
centimeters, and πΊ is its center of mass. Since weβre looking for the center
of mass of this resulting shape and weβre explicitly told that it was an equilateral
triangle with πΊπ΅πΆ cut out of it, this suggests that we should use the negative
mass method. To use the negative mass method, we
first start with this diagram here that shows the equilateral triangle π΄π΅πΆ shaded
in orange and the triangle πΊπ΅πΆ shaded over it in magenta.

Separating this out into its two
pieces, we have the original lamina π΄π΅πΆ and the lamina π΅πΊπΆ that represents the
part that was cut out. The uniform lamina π΄π΅πΆ is a
regular object, so its mass is positive. However, since the combination of
πΊπ΅πΆ and π΄π΅πΆ results in a lamina with no mass at the bottom, where π΄π΅πΆ
clearly does have mass, πΊπ΅πΆ must have negative mass because negative plus
positive is zero. We find the resulting center of
mass the way we would find the center of mass of any two laminae. Itβs just one of these laminae now
has a negative number instead of a positive number for its mass. This is why we call it the negative
mass method.

We now need to find the mass and
the location of the center of mass of these two laminae. Since these laminae are uniform, we
can express their total mass as their area times a constant, which is the mass per
unit area of the material. For the lamina with positive mass,
we need the area of the equilateral triangle π΄π΅πΆ. From the diagram, we know the
length of the base is 93 centimeters and the height is the length of the blue dashed
line, which passes through the points π΄ and πΊ. Because this dashed line is an
altitude, it is perpendicular to the line π΅πΆ. Furthermore, because π΄π΅πΆ is
equilateral, this angle here at point π΅ is 60 degrees. So the triangle formed by π΄π΅ and
the point where the altitude meets the side π΅πΆ is a 30-60-90 right triangle.

Either directly using trigonometry
or using facts about 30-60-90 right triangles, since the hypotenuse of this triangle
is 93 centimeters, the height must be 93 times the square root of three divided by
two. Weβll actually calculate a definite
value for this later, but for now, letβs just call it π. So the triangle π΄π΅πΆ has a base
of 93 and a height of π. So its area, one-half base times
height, is one-half 93π. If we call the constant uniform
density π, then one-half times 93 times π is the area of π΄π΅πΆ, and one-half 93π
times π is the total mass of the lamina π΄π΅πΆ.

What about πΊπ΅πΆ? Well, recall that πΊ is the center
of mass of the equilateral triangle. For any uniform triangle, the
center of mass is one-third of the way along any of the medians. In any collateral triangle, the
median which connects any vertex to the midpoint of the opposite side is also the
altitude of the triangle. So the distance from the midpoint
of π΅πΆ to the point πΊ is one-third of the total height of the triangle π΄π΅πΆ. But we know what that value is; we
called it π. So the height of π΅πΊπΆ is
one-third of π. π΅πΊπΆ and π΄π΅πΆ have the same
base. So the area of π΅πΊπΆ is one-half
times 93 times one-third of π. Now remember, π΅πΊπΆ needs to have
negative mass but have the same kind of mass distribution as π΄π΅πΆ. So the density of π΅πΊπΆ will be
negative π.

Letβs rearrange our expression for
the mass of π΅πΊπΆ by factoring out the negative one and the factor of
one-third. Written this way, we see that the
mass of π΅πΊπΆ is negative one-third times the mass of π΄π΅πΆ. The factor of negative one-third is
present because π΄π΅πΆ has three times the area of π΅πΊπΆ, and the mass of π΅πΊπΆ is
negative. Just to make our life easier and
clean up some space, letβs call one-half times 93 times π times lowercase π
capital π so that the mass of π΄π΅πΆ is capital π and the mass of π΅πΊπΆ is
negative one-third times capital π.

Now that we know the masses, letβs
find the centers of mass of π΄π΅πΆ and π΅πΊπΆ. We know that the center of mass of
π΄π΅πΆ is located at the point πΊ. And weβve already determined the
π¦-coordinate of point πΊ; itβs one-third of π. πΊ lies along the altitude of
π΄π΅πΆ. And we know that this altitude also
being a median intersects the π₯-axis halfway between π΅ and πΆ. Well, since the length of π΅πΆ is
93, halfway between them is 46.5. So the coordinates of point πΊ are
46.5 comma one-third π. For triangle π΅πΊπΆ, note that
π΅πΊπΆ is an isosceles triangle. We can see this because the dashed
line is the perpendicular bisector of π΅πΆ. So any point on this bisector
including the point πΊ is equidistant from point π΅ and point πΆ. So the sides π΅πΊ and πΊπΆ are
congruent, and therefore π΅πΊπΆ is isosceles.

In an isosceles triangle, the
altitude drawn from the vertex in between the two congruent sides is also a median,
just like in the equilateral triangle. But that means that the altitude of
π΅πΊπΆ, which is the dotted line that weβve already drawn, is also a median. Since the center of mass is exactly
one-third of the way along the median, the center of mass of π΅πΊπΆ is right around
here. Although weβve only estimated the
location of this point in the figure, we can still find its location exactly. Itβs on the same vertical line as
πΊ, so its π₯-coordinate is also 46.5. The π¦-coordinate is one-third of
the height of π΅πΊπΆ, and the height of π΅πΊπΆ is one-third of π. So the π¦-coordinate is one-third
of one-third or one-ninth of π.

Now we have the mass and center of
mass of π΅πΊπΆ and π΄π΅πΆ. To use this information to find
what weβre looking for, we can use the center of mass formula to find each
coordinate of the resulting center of mass of the combination of π΄π΅πΆ and
π΅πΊπΆ. For each coordinate, we plug in the
mass and corresponding coordinate of the center of mass for each of our laminae into
the numerator and the mass of each lamina into the denominator. Letβs start with the
π¦-coordinate. The mass of π΄π΅πΆ is π, and the
π¦-coordinate of the center of mass is one-third of π. So the first term of the sum of our
numerator is π times one-third π. The second term in the numerator is
negative one-third π times one-ninth π, which is the mass of π΅πΊπΆ times the
π¦-coordinate of its center of mass.

The denominator is the sum of the
two masses, π plus negative one-third π. To start simplifying this fraction,
letβs note that both terms in the numerator have an π, a one-third, and an π. So weβll factor one-third times π
times π out of both of these terms. In the denominator, π plus
negative one-third π is just two-thirds π. So we have one-third times π times
π times one minus one-ninth all divided by two-thirds π. π divided by π is one, and
one-third in the numerator divided by two-thirds in the denominator leaves just two
in the denominator. In the numerator, one minus
one-ninth is eight-ninths. So we have π times eight-ninths
divided by two.

Note that the actual value of the
mass of π΄π΅πΆ does not factor into this calculation because the only thing thatβs
important is the relative mass of π΄π΅πΆ and π΅πΊπΆ not their absolute masses. Eight-ninths divided by two is
four-ninths. So π times eight-ninths divided by
two is four-ninths times π. Plugging back in our definition for
π, we get that the π¦-coordinate of the center of mass of our new lamina is
four-ninths times 93 times the square root of three divided by two. If we plug this into a calculator,
we get 35.7957 and several more decimal places. Since weβre asked to round to two
decimal places, the third decimal place is a five. So we add one to the second decimal
place, which is nine. One plus nine is 10, so 35.7957 et
cetera rounds to 35.80.

To find the π₯-coordinate of the
center of mass, instead of going through this whole calculation again, we note that
the π₯-coordinate of the center of mass for π΄π΅πΆ and π΅πΊπΆ is the same. This means that when we combine
them, we wonβt change the π₯-coordinate of the center of mass. So the π₯-coordinate of the center
of mass of our resulting lamina will also be 46.5. So rounding as specified, we find
that the center of mass of our new lamina is at 46.5 comma 35.80.