Difference between revisions of "2009 AMC 12A Problems/Problem 10"
(New page: == Problem == In quadrilateral <math>ABCD</math>, <math>AB = 5</math>, <math>BC = 17</math>, <math>CD = 5</math>, <math>DA = 9</math>, and <math>BD</math> is an integer. What is <math>BD</...) |
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| + | {{duplicate|[[2009 AMC 12A Problems|2009 AMC 12A #10]] and [[2009 AMC 10A Problems|2009 AMC 10A #12]]}} | ||
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== Problem == | == Problem == | ||
In quadrilateral <math>ABCD</math>, <math>AB = 5</math>, <math>BC = 17</math>, <math>CD = 5</math>, <math>DA = 9</math>, and <math>BD</math> is an integer. What is <math>BD</math>? | In quadrilateral <math>ABCD</math>, <math>AB = 5</math>, <math>BC = 17</math>, <math>CD = 5</math>, <math>DA = 9</math>, and <math>BD</math> is an integer. What is <math>BD</math>? | ||
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label("$A$",A,NE); | label("$A$",A,NE); | ||
</asy></center><math>\textbf{(A)}\ 11 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 13 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 15</math> | </asy></center><math>\textbf{(A)}\ 11 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 13 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 15</math> | ||
| + | [[Category: Introductory Geometry Problems]] | ||
== Solution == | == Solution == | ||
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By the [[triangle inequality]] we have <math>BD < DA + AB = 9 + 5 = 14</math>, and also <math>BD + CD > BC</math>, hence <math>BD > BC - CD = 17 - 5 = 12</math>. | By the [[triangle inequality]] we have <math>BD < DA + AB = 9 + 5 = 14</math>, and also <math>BD + CD > BC</math>, hence <math>BD > BC - CD = 17 - 5 = 12</math>. | ||
| − | We | + | We get that <math>12 < BD < 14</math>, and as we know that <math>BD</math> is an integer, we must have <math>BD=\boxed{13}</math>. |
== See Also == | == See Also == | ||
{{AMC12 box|year=2009|ab=A|num-b=9|num-a=11}} | {{AMC12 box|year=2009|ab=A|num-b=9|num-a=11}} | ||
| + | {{AMC10 box|year=2009|ab=A|num-b=11|num-a=13}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 08:31, 8 June 2021
- The following problem is from both the 2009 AMC 12A #10 and 2009 AMC 10A #12, so both problems redirect to this page.
Problem
In quadrilateral 
, 
, 
, 
, 
, and 
is an integer. What is ?
![[asy] unitsize(4mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=4; pair C=(0,0), B=(17,0); pair D=intersectionpoints(Circle(C,5),Circle(B,13))[0]; pair A=intersectionpoints(Circle(D,9),Circle(B,5))[0]; pair[] dotted={A,B,C,D}; draw(D--A--B--C--D--B); dot(dotted); label("$D$",D,NW); label("$C$",C,W); label("$B$",B,E); label("$A$",A,NE); [/asy]](http://latex.artofproblemsolving.com/5/7/0/570b698fea7c6d66beab04dd6aee44dd24f2b6c1.png)
Solution
By the triangle inequality we have 
, and also 
, hence 
.
We get that 
, and as we know that is an integer, we must have 
.
See Also
| 2009 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 9 |
Followed by Problem 11 |
| 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 | |
| 2009 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
