Difference between revisions of "2008 iTest Problems/Problem 99"
(solution) |
(→Problem) |
||
(One intermediate revision by one other user not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | Given a convex, <math>n</math>-sided polygon <math>P</math>, form a <math>2n</math>-sided polygon <math>\text{clip}(P)</math> by cutting off each corner of <math>P</math> at the edges’ trisection points. In other words, <math>\text{clip}(P)</math> is the polygon whose vertices are the <math>2n</math> edge trisection points of <math>P</math>, connected in order around the boundary of <math>P</math>. Let <math>P_1</math> be an isosceles trapezoid with side lengths <math>13, 13, 13</math>, and <math>3</math>, and for each <math>i | + | Given a convex, <math>n</math>-sided polygon <math>P</math>, form a <math>2n</math>-sided polygon <math>\text{clip}(P)</math> by cutting off each corner of <math>P</math> at the edges’ trisection points. In other words, <math>\text{clip}(P)</math> is the polygon whose vertices are the <math>2n</math> edge trisection points of <math>P</math>, connected in order around the boundary of <math>P</math>. Let <math>P_1</math> be an isosceles trapezoid with side lengths <math>13, 13, 13</math>, and <math>3</math>, and for each <math>i \geq 2</math>, let <math>P_i = \text{clip}(P_{i-1})</math>. This iterative clipping process approaches a limiting shape <math>P_\infty = \lim_{i \rightarrow \infty} P_i</math>. If the difference of the areas of <math>P_{10}</math> and <math>P_{\infty}</math> is written as a fraction <math>\frac{x}{y}</math> in lowest terms, calculate the number of positive integer factors of <math>x \cdot y</math>. |
== Solution == | == Solution == | ||
Line 11: | Line 11: | ||
Let <math>A_1, A_2</math> be the trisection points on <math>\overline{AB},\overline{AD}</math>, respectively, that are closest to <math>A</math>. Then the operation <math>\text{clip}(P)</math> deletes <math>\triangle A_1AA_2</math>. Since <math>A_1A/AB = 1/3, A_2A/AD = 1/3</math>, and <math>\triangle A_1AA_2, \triangle BAD</math> share common <math>\angle A</math>, we have <math>\triangle A_1AA_2 \sim \triangle BAD</math> by side ratio <math>1/3</math>. Their areas are in the ratio <math>(1/3)^2 = 1/9</math>. | Let <math>A_1, A_2</math> be the trisection points on <math>\overline{AB},\overline{AD}</math>, respectively, that are closest to <math>A</math>. Then the operation <math>\text{clip}(P)</math> deletes <math>\triangle A_1AA_2</math>. Since <math>A_1A/AB = 1/3, A_2A/AD = 1/3</math>, and <math>\triangle A_1AA_2, \triangle BAD</math> share common <math>\angle A</math>, we have <math>\triangle A_1AA_2 \sim \triangle BAD</math> by side ratio <math>1/3</math>. Their areas are in the ratio <math>(1/3)^2 = 1/9</math>. | ||
− | + | Similarly, <math>[C_1CC_2] = \frac{1}{9}[BCD]</math>, and <math>[A_1AA_2] + [C_1CC_2] = \frac{1}{9}[ABCD]</math>. Cutting along diagonal <math>AC</math>, we get the same result, so <math>D_1 = \frac{2}{9}P_1</math>. | |
Latest revision as of 13:39, 7 October 2017
Problem
Given a convex, -sided polygon , form a -sided polygon by cutting off each corner of at the edges’ trisection points. In other words, is the polygon whose vertices are the edge trisection points of , connected in order around the boundary of . Let be an isosceles trapezoid with side lengths , and , and for each , let . This iterative clipping process approaches a limiting shape . If the difference of the areas of and is written as a fraction in lowest terms, calculate the number of positive integer factors of .
Solution
Let be the difference in the areas between and . Let our trapezoid be (and ); then without loss of generality construct diagonal .
Let be the trisection points on , respectively, that are closest to . Then the operation deletes . Since , and share common , we have by side ratio . Their areas are in the ratio .
Similarly, , and . Cutting along diagonal , we get the same result, so .
We now consider the effects of the second clipping. Without loss of generality consider what happens along the vertex of . Let be the trisection point along (again closest to ), and be the trisection point along . Now and , and . Using the definition of the area of a triangle, we see that . A similar clipping about gives ; around each clipped region in , we clip a new area . Generalizing, we have the recursion .
Then, . Hence,
Then has factors.