Difference between revisions of "2017 UNCO Math Contest II Problems/Problem 6"
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== Problem == | == Problem == | ||
| + | <asy> | ||
| + | pair A=dir(72),B=dir(144),C=dir(216),D=dir(288),E=dir(360),O=(0,0); | ||
| + | draw(A--B--C--D--E--A); | ||
| + | pair AB1=(A+2*B)/3,AB2=(A+B)/2,AB3=(2*A+B)/3; | ||
| + | draw(C--AB1--C--AB2--C--AB3); | ||
| + | pair BC1=(B+2*C)/3,BC2=(B+C)/2,BC3=(2*B+C)/3; | ||
| + | draw(D--BC1--D--BC2--D--BC3); | ||
| + | pair CD1=(C+2*D)/3,CD2=(C+D)/2,CD3=(2*C+D)/3; | ||
| + | draw(E--CD1--E--CD2--E--CD3); | ||
| + | pair DE1=(D+2*E)/3,DE2=(D+E)/2,DE3=(2*D+E)/3; | ||
| + | draw(A--DE1--A--DE2--A--DE3); | ||
| + | pair EA1=(E+2*A)/3,EA2=(E+A)/2,EA3=(2*E+A)/3; | ||
| + | draw(B--EA1--B--EA2--B--EA3); | ||
| + | </asy> | ||
| + | |||
| + | The Spider's Divider | ||
| + | |||
| + | On a regular pentagon, a spider | ||
| + | forms segments that connect one endpoint of each | ||
| + | side to n different non-vertex points on the side adjacent | ||
| + | to the other endpoint of that side, going around | ||
| + | clockwise, as shown. Into how many non-overlapping | ||
| + | regions do the segments divide the pentagon? Your answer | ||
| + | should be a formula involving n. (In the diagram, | ||
| + | n = 3 and the pentagon is divided into 61 regions.) | ||
== Solution == | == Solution == | ||
| + | We can see that the web is made out of 5 [[congruent]] regions surrounding one central region. to find the number of parts in one of the five congruent regions, which is basically <math>n</math> sections each divided by <math>n</math> lines, so each section is divided into <math>n+1</math> parts. The formula is thus <math>5n(n+1)+1</math> or <math>\boxed{5n^2+5n+1}</math> | ||
== See also == | == See also == | ||
{{UNCO Math Contest box|year=2017|n=II|num-b=5|num-a=7}} | {{UNCO Math Contest box|year=2017|n=II|num-b=5|num-a=7}} | ||
| − | [[Category:Intermediate | + | [[Category:Intermediate Geometry Problems]] |
Latest revision as of 16:53, 16 January 2023
Problem
![[asy] pair A=dir(72),B=dir(144),C=dir(216),D=dir(288),E=dir(360),O=(0,0); draw(A--B--C--D--E--A); pair AB1=(A+2*B)/3,AB2=(A+B)/2,AB3=(2*A+B)/3; draw(C--AB1--C--AB2--C--AB3); pair BC1=(B+2*C)/3,BC2=(B+C)/2,BC3=(2*B+C)/3; draw(D--BC1--D--BC2--D--BC3); pair CD1=(C+2*D)/3,CD2=(C+D)/2,CD3=(2*C+D)/3; draw(E--CD1--E--CD2--E--CD3); pair DE1=(D+2*E)/3,DE2=(D+E)/2,DE3=(2*D+E)/3; draw(A--DE1--A--DE2--A--DE3); pair EA1=(E+2*A)/3,EA2=(E+A)/2,EA3=(2*E+A)/3; draw(B--EA1--B--EA2--B--EA3); [/asy]](http://latex.artofproblemsolving.com/e/0/e/e0e9809b75e06e64e84ac5f3f586c39872eeec32.png)
The Spider's Divider
On a regular pentagon, a spider forms segments that connect one endpoint of each side to n different non-vertex points on the side adjacent to the other endpoint of that side, going around clockwise, as shown. Into how many non-overlapping regions do the segments divide the pentagon? Your answer should be a formula involving n. (In the diagram, n = 3 and the pentagon is divided into 61 regions.)
Solution
We can see that the web is made out of 5 congruent regions surrounding one central region. to find the number of parts in one of the five congruent regions, which is basically 
sections each divided by lines, so each section is divided into 
parts. The formula is thus 
or 
See also
| 2017 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
| All UNCO Math Contest Problems and Solutions | ||
