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Question Number 21097 by Tinkutara last updated on 12/Sep/17

Suppose in the plane 10 pairwise nonparallel lines intersect one another. What is the maximum possible number of polygons (with finite areas) that can be formed?

$$\mathrm{Suppose}\:\mathrm{in}\:\mathrm{the}\:\mathrm{plane}\:\mathrm{10}\:\mathrm{pairwise} \\ $$$$\mathrm{nonparallel}\:\mathrm{lines}\:\mathrm{intersect}\:\mathrm{one}\:\mathrm{another}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{possible}\:\mathrm{number} \\ $$$$\mathrm{of}\:\mathrm{polygons}\:\left(\mathrm{with}\:\mathrm{finite}\:\mathrm{areas}\right)\:\mathrm{that}\:\mathrm{can} \\ $$$$\mathrm{be}\:\mathrm{formed}? \\ $$

Answered by Tinkutara last updated on 15/Sep/17

Let us assume a_n represents the number of regions of plane formed by pairwise nonparallel lines. Then T_1 =2,T_2 =4,T_3 =7,T_4 =11 and so on. Let S=2+4+7+11+...+T_n S= 2+4+7+11+...+T_n 0=2+[2+3+4+...(n−1)terms]−T_n T_n =1+((n(n+1))/2) ∴ T_(10) =1+((10×11)/2)=56 Seeing the number of nonoverlapping polygons formed by upto 4 lines, we get number of polygons, N=T_n −2n. ∴ Number of polygons formed by 10 nonparallel lines = 56−20=36.

$${Let}\:{us}\:{assume}\:{a}_{{n}} \:{represents}\:{the} \\ $$$${number}\:{of}\:{regions}\:{of}\:{plane}\:{formed} \\ $$$${by}\:{pairwise}\:{nonparallel}\:{lines}. \\ $$$${Then}\:{T}_{\mathrm{1}} =\mathrm{2},{T}_{\mathrm{2}} =\mathrm{4},{T}_{\mathrm{3}} =\mathrm{7},{T}_{\mathrm{4}} =\mathrm{11}\:{and}\:{so}\:{on}. \\ $$$${Let}\:{S}=\mathrm{2}+\mathrm{4}+\mathrm{7}+\mathrm{11}+...+{T}_{{n}} \\ $$$$\:\:\:\:\:\:\:{S}=\:\:\:\:\:\:\:\:\mathrm{2}+\mathrm{4}+\mathrm{7}+\mathrm{11}+...+{T}_{{n}} \\ $$$$\mathrm{0}=\mathrm{2}+\left[\mathrm{2}+\mathrm{3}+\mathrm{4}+...\left({n}−\mathrm{1}\right){terms}\right]−{T}_{{n}} \\ $$$${T}_{{n}} =\mathrm{1}+\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\therefore\:{T}_{\mathrm{10}} =\mathrm{1}+\frac{\mathrm{10}×\mathrm{11}}{\mathrm{2}}=\mathrm{56} \\ $$$${Seeing}\:{the}\:{number}\:{of}\:{nonoverlapping} \\ $$$${polygons}\:{formed}\:{by}\:{upto}\:\mathrm{4}\:{lines}, \\ $$$${we}\:{get}\:{number}\:{of}\:{polygons},\:{N}={T}_{{n}} −\mathrm{2}{n}. \\ $$$$\therefore\:{Number}\:{of}\:{polygons}\:{formed}\:{by} \\ $$$$\mathrm{10}\:{nonparallel}\:{lines}\:=\:\mathrm{56}−\mathrm{20}=\mathrm{36}. \\ $$

Commented by Tinkutara last updated on 15/Sep/17

Question is exactly the same as it appeared in PRMO 2017. I think somewhere is miswording in question. But since all answers are 2-digit integers in this test, so 36 can be a possibility.

$$\mathrm{Question}\:\mathrm{is}\:\mathrm{exactly}\:\mathrm{the}\:\mathrm{same}\:\mathrm{as}\:\mathrm{it} \\ $$$$\mathrm{appeared}\:\mathrm{in}\:\mathrm{PRMO}\:\mathrm{2017}.\:\mathrm{I}\:\mathrm{think} \\ $$$$\mathrm{somewhere}\:\mathrm{is}\:\mathrm{miswording}\:\mathrm{in}\:\mathrm{question}. \\ $$$$\mathrm{But}\:\mathrm{since}\:\mathrm{all}\:\mathrm{answers}\:\mathrm{are}\:\mathrm{2}-\mathrm{digit}\:\mathrm{integers} \\ $$$$\mathrm{in}\:\mathrm{this}\:\mathrm{test},\:\mathrm{so}\:\mathrm{36}\:\mathrm{can}\:\mathrm{be}\:\mathrm{a}\:\mathrm{possibility}. \\ $$

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