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Topic: POW solution, Feb. 28-March 4
Replies: 0

 Problem of the Week Posts: 292 Registered: 12/3/04
POW solution, Feb. 28-March 4
Posted: Mar 7, 1994 10:59 PM

Many good answers came in for last week's problem.

************************************************
Problem of the Week, February 28 to March 4

o
o Jill and Jimmy were on their lunch break from the
O bridge building assignment. Jimmy was blowing soap
_ bubbles to pass the time. One of Jimmy's bubbles
' ` _ turned out to look a little like the one at the
left.
/ o \ ' `
| \ Suppose Jimmy's "double bubble" consists of two
\ | spheres whose centers are 14 cm apart. If the two
` _ '\ spheres intersect in a circle whose radius is 12
cm,
` o / and the radius of one sphere is 13 cm, what is the
` _ ' radius of the other sphere?
o
************************************************

The following all submitted accurate solutions:

Christina Molinari Grade 9 Masterman Phila.
Hannah Guhm Grade 9 Masterman Phila.
Percy Rosales Grade 9 Masterman Phila.
Derek Morrison and Jill Grobelski, Grade 9, Steel Valley High School, Pa.
Joe Sanchez Grade 10, Shaler Area HIgh School
Susie Taylor Grade 10, Shaler Area High School
Jennifer Strong Grade 10, Shaler Area High School
Nick Szmyd Grade 10, Shaler Area High School
Tim Thiel Grade 10, Shaler Area High School
Jerry Shanko Grade 10, Shaler Area High School
Ian Ross Grade 10, Shaler High School
Rachel Laughlin Grade 10, Shaler Area H.S.
Jim Michalek Grade 10, Shaler Area High School
Tony Aiello Grade 10, Shaler Area High School
Mike Gelik Grade 10, Shaler Area High School
Leslie Shay Grade 10, Burnaby South Sec School, Burnaby BC
Daniel Chan Grade 10, Burnaby South Sec School, Burnaby BC

Below are examples of correct solutions.

******************

From: strong jenn <strong@one.sasd.k12.pa.us>

My name is Jennifer Strong and I am in a tenth grade honors geometry
class at Shaler Area High School.

It is given that that the segment joining the centers of the spheres
is 14 units long. This segment would be perpendicular to the circle
created by the joining of the bubbles which has a radius of 12 units.
Two right trianlges are formed when the radii of the spheres are
drawn so that one endpoint is at the center of the sphere, and the
other endpoint is on the outer egde of the "Joint" circle. (For
convenience, let's assume that the radii meet at the same point at
the very top of the "Joint" circle). The first triangle formed would
have sides of 13 (the radius of the given sphere) which will be "c"
and 12 (the radius of the "joint" circle) which will be "a". The
other side can be represtented with the variable "b". Using the
Pythagorean theorem, (a^2+b^2=c^2) and switching it around (addition
property of equality) b^2=c^2-a^2. Substituting the given values,
b^2= 13^2 - 12^2. b^2 = 25 b=5.
That is, 5 is the distance (perpendicular distance) from the center
of the sphere with a radius of 13 to the center of the circle. Also
knowing that the total distance from the center of one sphere to the
center of the other sphere is 14, we can easily subtract (14-5) to
find the distance from the center of the other sphere to the center
of the "joint" circle. This distance is found to be 9. Again using
the Pythagorean Theorem, and the knowledge that the radius of the
"joint" circle is 12, we can find the other side of the triangle,
which would be the distance from the center of the sphere to the top
of the circle, which is also its radius. 9^2+12^2 = radius of the
sphere ^2 = 225. The square root of 225 is 15. So, 15 is the

*********************

From: nick szmyd <szmyd@one.sasd.k12.pa.us>

In starting this problem, I drew a picture of the "double
bubble". I plotted the center of each bubble A and C. I then plotted
the points where the intersection began and ended. The top
intersection, point B. The bottom intersection, point D. I then
connected all of the points to form a quadrilateral. I then drew
segments BD and AC and called the intersection E. As the distances
given in the problem, AC=14 cm. BE=12 cm and ED=12 cm since they are
both radii of the circle intersection. Bc and DC each equal 13 cm
since they are both the radius of one bubble. In assuming that
angles AEB, BEC, CED, and DEA all are right angles, I used
a^2+b^2=c^2 to figure out the other sides of the triangles. I
calculated that the side EC was 5 cm long. Since the distance between
the centers is 14 cm and the distance from the center of one bubble
to the intersection is 5 cm, I did 14-5 to equal the distance from
the other bubbles center to the intersection, or 9 cm. So EA=9 cm. I
again used a^2+b^2=c^2 to figure out side AB and AD and figured out
the side's length is 15 cm long which is the radius of the other
bubble.

************************

From: tim thiel <thiel@one.sasd.k12.pa.us>

(please double click on the icon to get my drawing of the
intersecting spheres)

Two spheres are drawn and connected by segment ab at their centers.
Points m and n were drawn on the intersection of the spheres which is
a circle. segments an,bn,am, and bm, are introduced into the picture
by the line postulate. from theorem it is found that the center of
the circle formed by the intersection is the foot of the
perpendicular from one of the intersecting spheres. Triangle mcb is
a right triangle because of the perpendicular line creating a right
angle(mcb). The radius of the circle created is given as 12 cm so
segments cm and cn are 12cm long. to find side cb:
cb=(mb^2-cm^2)^.5
cb=(13^2-12^2)^.5
cb=(25)^.5
cb=5cm (used the pythagoreon theorem)
side ac=ab-cb
side ac=9cm
to find side am which is the radius of s1 use the pythagoreon
theorem on the right triangle mca:
am=(mc^2+ac^2)^.5
am=(12^2+9^2)^.5
am=(225)^.5
am=15 cm
*****The radius of the other sphere(in this case sphere 1 is found to
be 15cm.*****

***********************

From: jerry shanko <shanko@one.sasd.k12.pa.us>

Let P and P' be the centers of the two spheres S and S'. A and B
are two points of intersection of the two spheres. Segment AB is a
diameter of the circle made by the spheres' intersection. Line PP'
is the perpendicular bisector of segment AB and intersects AB at
point M, which is the center of the circle. Line PA is tangent to S'
at A. Triangle PAM is a right triangle with PA=13cm and AM=12cm.
Using the Pythagorean Theorem, segment PMis found to be 5cm . Since
PP'=14cm and PM=5cm, then subtracting you find MP'=9cm. Using the
Pythagorean Theorem again, AP' is found to be 15cm. The radius of
the second sphere is 15cm.

Drawing included.

*********************
From: m1 <m1@one.sasd.k12.pa.us>

Hi, this is Ian Ross grade 10 Shaler High School

The radius of the other triangle is 15 cm. Here is how I solved it.

Apply the pythagorean theorem. a^2+b^2=c^2
Assume the two bubbles are of different sizes.

12^2+r^2=13^2
144+r^2=169
r^2=25
r=5
14-5=9
9^2+12^2=r^2
81+144=r^2
225=r^2
r=15