Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
Drexel University or The Math Forum.
|
|
|
|
Geometry POW Solution, February 3-7
Posted:
Feb 22, 1997 12:42 PM
|
|
*************************************************
Geometry Problem of the Week, February 3-7
I was going to do a problem about Groundhog Day but wasn't having very
much success. Groundhogs are not all that geometric. Instead I will use
a problem that Jon Sheffi sent me. Jon is in 10th grade at the Commonwealth
School in Boston, and this problem appeared on his geometry mid-term. I
hope his teacher doesn't mind me showing it to the whole world! (And
yes, Jon better get the problem right this week!)
I had fun doing this problem, so I thought it might make a good one for
y'all. You're gonna need to draw a good picture, that's for sure!
AOD is a diameter of circle O. B is any point on the circle. At B, a
tangent is drawn to the circle. From the center, O, a line is drawn parallel
to AB, meeting the tangent at P. Prove that PD is tangent to the circle.
*************************************************
This was a neat problem that Jon Sheffi sent me. And I learned a lot
reading the solutions - like the fact that I didn't do a very good job of
figuring it out myself. Though I was right, there are several ways to do
it that are slicker than mine. In fact, nobody else solved it the way I
did, so that shows that it was a bit weird!
29 people got this right, and 15 got it wrong. We also welcome five new
schools this week, Mills High School from Millbrae, California,
Cistercian Prep from Palo Alto, California, Cloverdale High School from
Santa Rosa, California, and our first schools from Italy and Senegal -
Salvo D'Acquisto School from Bologna, Italy, and Institution Sainte
Jeanne D'Arc from Dakar, Senegal. Pretty exciting!
The basic idea to everyone's solution was the same - prove that those two
triangles are congruent so that you can show that ODP is a right angle.
There were a lot of ways to do this. My way, which involved a lot of
angle-addition stuff and NO parallel line stuff, proved to be unpopular.
Brent Tworetsky and a coupld of other folks used exterior angles of
triangles and the sum of remote interior angles. A bunch of people
incorporated vertical angles and added a lot of segments. Phong Pham of
Highland Park Senior High School and Thomas Kuo of Burroughs High School
used the fact that parallel lines through a circle will cut out equal
arcs, and hence equal angles. Thomas' proof is included below.
YA from Smoky Hill High School wrote a nice two-columm
proof. Danny Ornstein and Rachel Winnik from Georgetown Day School wrote
the shortest possible proof, and provided a very nice introductory
statement about what they wanted to do and how they were going to do it.
Both of these proofs are shown below for your reading pleasure.
Mahdi Tarraf from the Institution Sainte Jeanne D'Arc wrote a nice
explanation that includes some fact and theorems about kites (without
actually calling them kites). This solution is below as well.
As often happens with proofs, a few people made unsupported statements
and then proved things from there. They used facts that were only true
if, say, PD WAS tangent to the circle, and then used those facts to prove
that PD IS tangent. It just doesn't work this way, and you need to be
careful when you're writing proofs that you don't do this!
Jason Lee of Stuyvesant High School tried to put one over on me by making
the assumption that three points were collinear and going from there. I
completely missed this, but he pointed it out to me when I told him I
thought it looked good. He should get points for correcting himself :-)
If you come across a problem on one of your tests or in your homework (or
anywhere!) that you think might be a good geometry POW, send it along!
***********************************************
The following students submitted correct solutions this week.
Highlighted solutions are included below. For a full list of solutions,
please check out
http://forum.swarthmore.edu/geopow/fullsolutions/020797.fullsolution.html.
Rick Peterson, Grade , David Sarnoff Research Center, Princeton, New Jersey
Brent Tworetzky, Grade 10, JP Taravella High School, Coral Springs, Florida
Christine Chung, Grade 10, Smoky Hill High School, Aurora, Colorado
YA, Grade 9, Smoky Hill High School, Aurora, Colorado
John Yang, Grade 12, Mills High School, Millbrae, California
Justin Lam, Grade 8, Sequoia Middle School, Pleasant Hill, California
Scott Devoe, Grade 9, Smoky Hill High School, Aurora, Colorado
Elizabeth Peterson, Grade 8, Lanier Middle School, Houston, Texas
Jon Sheffi, Grade 10, Commonwealth School, Boston, Massachusetts
Tana Kaplan and Marnie Hanel, Grade , Lakeside School, Seattle, Washington
Lauren Deal and Riely King and Andrew Jonsson and John Alexander, Grade ,
Lakeside School, Seattle, Washington
Nicole Giuliani, Grade 10, Lakeside School, Seattle, Washington
Phong Pham, Grade 12, Highland Park Senior High School, St. Paul, Minnesota
Gavin Calkins, Grade 8, Waluga Junior High School, Portland, Oregon
Kate Maxwell and Lauren Levien, Grade 8, Georgetown Day School, Washington, DC
Dave Peterson, Grade homeschool parent, Rochester, New York
Doug Dickson, Grade , Lakeside School, Seattle, Washington
Andrew Novion, Grade 10, Lakeside School, Seattle, Washington
Jenny Kaplan, Grade 6, Castilleja Middle School, Palo Alto, California
Brian Gordon, Grade Dartmouth '92, Wethersfield, Connecticut
Dirk Spoons, Grade 8, Murray Middle School, Ridgecrest, California
Shameica Edwards, Grade 11, George Wingate High School, Brooklyn, New York
Daniel Ornstein and Rachel Winnik, Grade 8, Georgetown Day School,
Washington, DC
Thomas Kuo, Grade 9, Burroughs High School, Ridgecrest, California
Jimmy Widger, Grade , Newport High School, Bellevue, Washington
Peter Liang, Grade 10, Newport High School, Bellevue, Washington
Janet Eckart, Grade 9, Granada High School, Livermore, California
11th grade class, Grade 11, Salvo D'Acquisto Secondary School, Bologna, Italy
Mahdi Tarraf, Grade 8, Institution Sainte Jeanne D'Arc, Dakar, Senegal
***********************************************
From: YA
anonymous@anonymous.com
Grade: 9
School: Smoky Hill High School, Aurora, Colorado
STATEMENTS REASONS
1. BO=OA 1. All radii of a circle are congruent.
2. triangle BOA is isosceles 2. If two sides of a triangle are congruent,
then it is a isosceles triangle.
3. <ABO=<BAO 3. If two sides of a triangle are congruent,
the angles opposite the sides are congruent.
4. <ABO=<POB 4. If parallel lines, then alternate interior
angles are congruent.
5. <POD=<BAO 5. If parallel lines, then corresponding angles
are congruent.
6. <BOP=<POD 6. Transitive property
7. BO=OD 7. All radii of a circle are congruent.
8. PO=PO 8. Reflexive property
9. triangles BOP and DOP 9. SAS
are congruent
10.<OBP=90 10.A tangent line is perpendicular to the
radius drawn to the point of contact.
11.<OBP=<PDO 11.CPCTC
12.<PDO=90 12.Transitive property
13.PD is a tangent 13.If a line is perpendicular to a radius at
its outer endpoint, then it is tangent
to the circle.
***********************************************
From: Daniel Ornstein and Rachel Winnik
gnielsen@triton.dmso.mil
Grade: 8
School: Georgetown Day School, Washington, DC
Subject: pow 2/7 from gds
Daniel Ornstein and
Rachel Winnik
Georgetown Day School
8th Grade
Washington DC
Paul Nass
<img src="http://forum.swarthmore.edu/geopow/gifs/020797.daniel.gif">
IN ORDER TO PROVE THAT LINE PD IS A TANGENT, WE MUST FIRST PROVE THAT ANGLE ODP
IS 90 DEGREES. IN ORDER TO DO THIS, WE CAN PROVE TRIANGLE OBP CONGRUENT TO
TRIANGLE ODP. THE PROOF FOR PROOVING THAT FOLLOWS:
STATEMENTS: REASONS:
--------------------------------|-----------------------------------
|
OP = OP | reflexive property
BO = DO = AO | all radii of the same circle are=
triangle AOB is isosceles | a triangle with 2 = sides is
| isosceles
angles 1 and 2 are = | base angles of an isoscles
| are congruent
angles 1=3, 2=4 | When two parallel lines are
| cut by a transversal, the
| alternate interior and
| corresponding angles are
| congruent
triangles OBP = ODP | side angle side
angle OBP = 90 degrees | Def. of a tangent
angle ODP =90 degrees | corresponding parts of
| = triangles are =.
***********************************************
From: Thomas Kuo
wkuo@ridgecrest.ca.us
Grade: 9
School: Burroughs High School, Ridgecrest, California
Subject: Solution of POW, Feb 3-7, 1997
From: Thomas Kuo
Email: wkuo@ridgecrest.ca.us
School: Sherman E. Burroughs High School, Ridgecrest, California
Grade: 9th
* F
*
* O
A * * * * * * * D
* *
* *
B * E
*
*
* P
AOD is the diameter of the circle O
line OP which is parallel to AB intersects the circle O
at E and F
Points A, B, E, D, and F are on the circle O
First we know that:
(a) PB is tangent to the circle at B makes angle OBP = 90 degrees
(b) angle AOF = angle DOP because AD interects FP
____________________________________________________________________________
(1) OB = OD because they are the radius of the circle
(2) arc AF = arc BE because AB || FP,
therefore, angle AOF = angle BOP = angle DOP
(3) OP = OP because of the reflexive property
(4) triangle BOP = triangle DOP by SAS property of congruence
(5) angle OBP = 90 degrees = angle ODP because of CPCTC
(6) PD is tangent to the circle at D because angle ODP = 90 degrees
AND THERE'S THE PROOF!!!
***********************************************
From: Mahdi Tarraf
tarraf@ns.arc.sn
Grade: 8
School: Institution Sainte Jeanne D'Arc, Dakar, Senegal
Subject: Geometry problem of the week
The triangle ABD is inscribed in the circle O which has for
diameter ad then by rule : '' If a triangle , is inscribed in a circle
and has for diameter one of his side, then this triangle is rectangle
and has for hypotenuse the diameter of this circle. Therefor ABD is a
rectangle triangle on B in this way, AB perpendicular to BD. And OP //
AB, then, by the rule : '' If two lines are parrallels, all
perpendiculars to one of them is perpendicular to the other therefore OP
perpendicular to BO BO = OD, then BOD is a triangle which has 2 sides
equals now, OP is a hight and in a triangle which has 2 sides equals the
hight is a perpendicular bisector too and in a triangle which has 2
equals side the perpendicular bisector is his axis' reflection too.
Therefore, the imageof the triangle OBP by the refection of axis OP is
PDO, but, this reflection keep the angles, the distance, therefore P^BO
= P^DO = 90ð then OD perpendicular to PD.
We have now D, contact point of the line (PD) and the circle and
PD perpendicular to the radius OD, but, all line perpendicular to a
radius on his contact's point on the circle is tangent to a circle.
Therefore PD is tangent to the circle O.
----------------
My name is Mahdi. I come from DAKAR (SENEGAL). I am 14 years
old. I pass my exam to finish the first cycle this year. I speak french
then excuse me for my langage errors.
I hope there will be a new problem soon.
MAHDI TARRAF.
***********************************************
|
|
|
|
