Counting Regions in the Plane and More 1

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1 Coutig Regios i the Plae ad More 1 by Zvezdelia Stakova Berkeley Math Circle Itermediate I Group September Overarchig Problem Problem 1 Regios i a Circle. The vertices of a polygos are arraged o a circle so that o three diagoals itersect i the same poit. How may regios iside the circle are formed this way by all the segmets coectig the poits? Problem Preparatio. Attackig Problem 1 for ay is hard. What is a good way to start o problems like this? What is your cojecture for the fial aswer?. Warm Up Problem 3 Warm-up. How may s are i the picture? Problem 4 Thikig Deeper How may differet ways ca be used to cout here? Which do you prefer? Which way is most geeralizable to other problems? 3. Special Vs. Geeric Positios Problem 5 Uderstadig the Coditios. Why did Problem 1 say that o three diagoals should itersect i a poit? If we allow for three or more segmets to itersect i a poit, will this chage the aswer? Will it icrease it? Decrease it? Defiitio 1 Geeral Positio. Two lies i the plae are said to be i geeral positio if they itersect i a poit. Three or more lie lies i the plae are said to be i geeral positio if ay two of them itersect, but o three or more lies itersect i the same poit. Problem 6 Special Cofiguratios. I the plae, whe shall we say that two lies are i a special positio? How about three lies i the plae i a special positio? Draw all differet cofiguratios of three lies i the plae i a special positio. Now draw all differet cofiguratios of three lies i the plae i a geeral positio. How may special ad how may geeric cofiguratios did you get? 1 Some problems ad pictures are take from A Decade of the Berkeley Math Circle The America Experiece, volume I, edited by Zvezdelia Stakova ad Tom Rike, published by the America Mathematical Society i the MSRI Mathematical Circles Library. 1

2 Problem 7 Petagos ad Hexagos. I a regular petago coect ay two vertices. Are the diagoals i special or i geeric positio? How about a regular hexago? Cout i each case the umber of regios ito which the polygo is cut up by its diagoals. 4. Solvig Predecessor Problems Problem 1 restricted us to lookig iside a circle? This might be why the problem is so hard! Let s look at the whole plae by temporarily elimiatig the circle. Problem 8 Lies i Geeric Positio. Draw lies i the plae so that o three itersect i the same poit ad o two are parallel. Ito how may regios do these lies divide the plae? Problem 9 Games i a Touramet. teams participated i a basketball touramet. Each team played every other team exactly oce. How may games were played i total? Problem 10 Coutig Diagoals. How may diagoals does the -go i our overarchig problem have? What is the relatio betwee the last three problems? 5. Pokig Aroud ad Extedig the Problems Problem 11 Debates i a Champioship. I a debate champioship with teams, each debate is doe betwee 3 teams. How may debates ca possible happe durig this touramet? How about if each debate ivolves 4 teams? Problem 1 Diagoal Itersectios. How may itersectios of diagoals does the -go i our overarchig problem have? What is the coectio betwee the last two problems? 6. The Fial Attack o the Overarchig Problem Problem 13 Addig a New Segmet. I the set-up of Problem 1, erase all segmets, but leave the poits ad the circle. Start all over the segmets oe at a time, i ay order. Recall that the segmets ca be either sides or diagoals of the -go. Suppose a ewly added segmet itersects k already draw diagoals. How may more regios iside the circle has this ew segmet added? Is there a elegat way to phrase the aswer without metioig the umber k? Problem 14 Puttig All Together. Use ay of our results so far to calculate the total umber of regios iside the circle. Is there a elegat way to phrase the aswer? Did you check it for correctess for 1,,..., 6? What does your aswer predict for 7? Check it by brute force, if you do t believe it.

3 7. Optioal Homework for the Die-Hards Problem 15 Summig Up. Prove the followig formulas for ay 1: 1 a 1 3 b c Problem 16 Off to Space. Geeralize the overarchig problem to 3 dimesios. The vertices of a polyhedro are arraged o a sphere so that whe amog the plaes coectig three of the poits, o three itersect i the same lie. How may regios iside the sphere are formed this way by all these plaes? Problem 17 Polygos. Cout the umber of regios iside a covex -go made by coectig ay two of its vertices if: a o three diagoals itersect at the same poit; b the -go is regular. Problem 18 Biomial Theorem. The umber of ways to choose k objects out of objects is deoted by the biomial coefficiet k read choose k. Prove that this biomial coefficiet is calculated by the formula:! for ay, k 0, k k! k! where 0! is defied to be 1 i order to make the formula work whe or k is 0. Whe < k or oe of the or k is < 0, the the biomial coefficiet k 0. Problem 19 Ultimate Summatios. Prove the followig formulas: a for all b 1 for odd Problem 0 Coicidece? The aswer 31 i the case of 6 poits i the overarchig Problem 1 broke the patter of powers k ad made us rethik what is goig o. The aswer to the warm-up Problem was also 31. Is this a coicidece, or is there a way to trasform oe problem ito the other i this case? For eve, the aalogous sum is a lot harder to calculate, but it turs out that parts of this particular sum is coected to our overarchig problem. 3

4 8. Short Aswers ad Ideas to Selected Problems Problem 1. Wait util the ed! No hurry. Keep thikig. Problem. Start with small s. For 1,, 3, 4, 5 we get 1,, 4, 8, 16 regios i the circle. A patter becomes apparet: powers of, i.e., 0, 1,, 3, ad 4. Thus, for 6 the umber of regios must be Yet, is it? Problem 3. Start with the biggest triagle ad keep addig segmets, coutig alog the way ad tryig to use the symmetry of the picture as much as possible: Problem 4. There are at least five possible ways of coutig the triagles. a Brute-force o system: Cout radomly ad hope that you did t miss ad you did t overcout aythig. b Icremetally: Recostruct the picture from scratch, addig icremetally simple parts such as segmets or small triagles ad coutig how may more figures of the wated type are added at each step. Start with the largest triagle, add up a segmet at a time ad cout how may ew triagles have bee created with that segmet as their side or part of their side. Start with a basic triagle, i.e., a triagle which is ot subdivided by aother segmet ito smaller triagles; add up a triagle at a time ad cout how may ew triagles have bee created that cotai your added triagle as part of them. d By umber of parts: Cout first the triagles that are made of oe piece, the those made of two pieces, the of three pieces, ad so o. e Symmetry: Use the cetral vertical lie as a symmetry divider ; cout triagles o oe side of it ad multiply by, the cout the triagles that go across the symmetry lie, ad the add. Did you get 31? 4

5 Problem 5. The umber of regios will decrease if we allow for three or more diagoals to itersect i a poit. For example, whe 6 see above, the umber of regios is 30, as the small cetral triagle has degeerated to the poit of itersectio of three diagoals. Problem 6. Two lies i the plae are i a special positio if they are parallel o poits of itersectio. There are 3 special cofiguratios of three lies i the plae: the three lies itersect i the same poit; or two lies are parallel ad the third lie i a trasversal itersects the other two lies; or all three lies are parallel. There is oly 1 geeric cofiguratio of three lies i the plae: a triagle whose sides have bee exteded to our three lies. Problem 7. The diagoals i a regular petago are i a geeric positio, but the diagoals i a regular hexago are i a special positio: three of them itersect i a poit. We have doe, i some sese, the coutig of the regios iside a petago ad iside a regular hexago already: subtract the rouded regios made by the circle i the overarchig problem discussio above. For a regular petago, you will get , ad for a regular hexago: Problem 8. Start with 0 lies dividig the plae ito 1 regio. If you already have draw k lies i geeral positio, addig a k 1st lie icreases the umber of regios by k 1 why?. Hece, overall for lies we get regios i the plae. Problem 9. Let s first solve the problem for five teams. PST 1. Arrage the teams i oe lie ad cout the umber of games, makig sure that you are ot overcoutig ay games. For 5 teams, this calculatio boils dow to Whe the problem is geeralized to teams, the result is 1 3 1, which requires a differet argumet to be calculated i a cocise form. PST. Cout the umber of games played by each team by double-coutig them oce for oe of the teams ad aother time for the other team, ad the divide by. For 5 teams, this yields agai 5 teams 4 games each Fortuately, this PST works for ay teams too: teams 1 games each As a result, we proved the followig:

6 Theorem 1. The sum of the first atural umbers is give by the formula Both sides of this formula couted differetly the same quatity: the umber of games to be played amog teams i a touramet. PST 3. To prove a formula LHS RHS, fid a quatity that equals both sides ad calculate it i two differet ways. I aother attempt to prove the above formula, draw a table to record the scores of all games. Alog the diagoal write X s, sice o team plays agaist itself. We are ot iterested i the outcomes of the games; so what is writte otherwise i the table does ot matter. What matters are the umber of o-diagoal cells: for every cell uder the diagoal there is a cell above the diagoal that refers to the same game, e.g., a cell for the game betwee teams i ad j ad a cell for the game betwee teams j ad i. Thus, the umber of games is the half of the o-diagoal cells: 1, where are the cells of the whole table, the subtracted are the diagoal cells, ad the divisio by offsets the double-coutig of the games. PST 4. To calculate a positive iteger quatity, fid a table whose cells approximate this quatity. Start overcoutig the quatity by the umber of cells i the table, ad the subtract ad divide as ecessary to elimiate uwated cells or cells that have bee couted multiple times. I a fial approach to solve the problem, we realize that the umber of games correspods to the umber of pairs of teams that we ca select from the give teams. There is a quatity i combiatorics which couts eve more geeral objects: these are the biomial coefficiets k from Problem 18. Applyig the Biomial Theorem for k yields agai!!! Problem 10. A diagoal rus betwee ay two vertices of polygo, except we have to subtract the sides of the polygo sice they cout as diagoals. From above, there are pairs of vertices, so the aswer is 1 6 3

7 Back to Problem 8, we ca ow give a ice formula for the umber of regios i the plae made by lies i geeric positio: Problem 11. ad Problem 1. Ay two itersectig diagoals form a covex quadrilateral, ad coversely, ay four poits o the circle form a covex quadrilateral whose diagoals itersect i a poit. Thus, the umber of diagoal itersectios equals the umber of 4-tuples of poits, i.e., 4. Problem 13. The ew segmet will icrease the umber of regios by k 1 why?. This works eve if the segmet is a side of the polygo, i.e., it has o diagoal itersectios it will still add ew regios. To record this elegatly, we will say that every diagoal itersectio is resposible for addig 1 ew regio, ad every segmet is resposible for addig 1 ew regio too. Problem 14. From Problem 13, the umber of regios equals 1 more tha the umber of diagoal itersectios plus the umber of segmets diagoals ad sides of the polygo, all couted. Problems 1 ad 10 give the aswers to the required two quatities: 4 ad. The 1 more comes from the fact that for 1 poit o the circle we start with 1 regio the whole circle, ad the every additioal segmet adds the amout of regios we discussed i Problem 13. Overall, the umber of regios iside the circle is: 1 for all For 1,, 3, 4, 5 this sum does equal to the correspodig power of, 1, but for 6 ad 7 we get , ad If you are familiar with fuctios ad, specifically, polyomials, the aswer ca be expressed as a degree 4 polyomial i : f a 3 b c 1, where we leave the iterested reader to calculate the coefficiets a, b, ad c. Problem You are o your ow. Oly, do t attempt these aloe at home: ask someoe for help ad view them as a log-term project. : 7