Difference between revisions of "2005 AMC 12A Problems/Problem 13"
Armalite46 (talk | contribs) m (→Solution) |
Franzliszt (talk | contribs) (→Solution 3) |
||
(10 intermediate revisions by 5 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | + | In the five-sided star shown, the letters <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math> and <math>E</math> are replaced by the | |
+ | numbers 3, 5, 6, 7 and 9, although not necessarily in that order. The sums of the | ||
+ | numbers at the ends of the line segments <math>\overline{AB}</math>, <math>\overline{BC}</math>, <math>\overline{CD}</math>, <math>\overline{DE}</math>, and <math>\overline{EA}</math> form an | ||
+ | arithmetic sequence, although not necessarily in that order. What is the middle | ||
+ | term of the arithmetic sequence? | ||
+ | |||
+ | <asy> | ||
+ | draw((0,0)--(0.5,1.54)--(1,0)--(-0.31,0.95)--(1.31,0.95)--cycle); | ||
+ | label("$A$",(0.5,1.54),N); | ||
+ | label("$B$",(1,0),SE); | ||
+ | label("$C$",(-0.31,0.95),W); | ||
+ | label("$D$",(1.31,0.95),E); | ||
+ | label("$E$",(0,0),SW); | ||
+ | </asy> | ||
<math> | <math> | ||
Line 6: | Line 19: | ||
</math> | </math> | ||
− | == Solution == | + | == Solutions == |
− | + | ||
− | === Solution | + | ===Solution 1=== |
+ | <math>(A+B) + (B+C) + (C+D) + (D+E) + (E+A) = 2(A+B+C+D+E)</math> (i.e., each number is counted twice). The sum <math>A + B + C + D + E</math> will always be <math>3 + 5 + 6 + 7 + 9 = 30</math>, so the arithmetic sequence has a sum of <math>2 \cdot 30 = 60</math>. The middle term must be the average of the five numbers, which is <math>\frac{60}{5} = 12 \Longrightarrow \mathrm{(D)}</math>. | ||
+ | ===Solution 2=== | ||
+ | Let the terms in the arithmetic sequence be <math>a</math>, <math>a + d</math>, <math>a + 2d</math>, <math>a + 3d</math>, and <math>a + 4d</math>. We seek the middle term <math>a + 2d</math>. | ||
+ | |||
+ | These five terms are <math>A + B</math>, <math>B + C</math>, <math>C + D</math>, <math>D + E</math>, and <math>E + A</math>, in some order. The numbers <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, and <math>E</math> are equal to 3, 5, 6, 7, and 9, in some order, so | ||
+ | <cmath>A + B + C + D + E = 3 + 5 + 6 + 7 + 9 = 30.</cmath> | ||
+ | Hence, the sum of the five terms is | ||
+ | <cmath>(A + B) + (B + C) + (C + D) + (D + E) + (E + A) = 2A + 2B + 2C + 2D + 2E = 60.</cmath> | ||
+ | But adding all five numbers, we also get <math>a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) = 5a + 10d</math>, so | ||
+ | <cmath>5a + 10d = 60.</cmath> | ||
+ | Dividing both sides by 5, we get <math>a + 2d = \boxed{12}</math>, which is the middle term. The answer is (D). | ||
+ | |||
+ | ===Solution 3=== | ||
+ | Not too bad with some logic and the awesome guess and check. Let <math>A=6</math>. Then let <math>B=7,E=5</math> and <math>C=3,D=9</math>. Our arithmetic sequence is <math>10,11,12,13,14</math> so our answer is <math>12 \Longrightarrow \mathrm{(D)}</math>. | ||
− | + | Solution by franzliszt | |
== See also == | == See also == |
Revision as of 18:52, 11 July 2020
Problem
In the five-sided star shown, the letters ,
,
,
and
are replaced by the
numbers 3, 5, 6, 7 and 9, although not necessarily in that order. The sums of the
numbers at the ends of the line segments
,
,
,
, and
form an
arithmetic sequence, although not necessarily in that order. What is the middle
term of the arithmetic sequence?
Solutions
Solution 1
(i.e., each number is counted twice). The sum
will always be
, so the arithmetic sequence has a sum of
. The middle term must be the average of the five numbers, which is
.
Solution 2
Let the terms in the arithmetic sequence be ,
,
,
, and
. We seek the middle term
.
These five terms are ,
,
,
, and
, in some order. The numbers
,
,
,
, and
are equal to 3, 5, 6, 7, and 9, in some order, so
Hence, the sum of the five terms is
But adding all five numbers, we also get
, so
Dividing both sides by 5, we get
, which is the middle term. The answer is (D).
Solution 3
Not too bad with some logic and the awesome guess and check. Let . Then let
and
. Our arithmetic sequence is
so our answer is
.
Solution by franzliszt
See also
2005 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.