************************************************* 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!
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 email@example.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 firstname.lastname@example.org 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
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 email@example.com Grade: 9 School: Burroughs High School, Ridgecrest, California
Subject: Solution of POW, Feb 3-7, 1997
From: Thomas Kuo Email: firstname.lastname@example.org 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
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 = 90ÃÂ° 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.