Difference between revisions of "2009 AMC 12A Problems/Problem 10"

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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>
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[[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 got 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>.
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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 ==
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{{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}}
 
{{AMC10 box|year=2009|ab=A|num-b=11|num-a=13}}
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{{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 $ABCD$, $AB = 5$, $BC = 17$, $CD = 5$, $DA = 9$, and $BD$ is an integer. What is $BD$?

[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]

$\textbf{(A)}\ 11 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 13 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 15$

Solution

By the triangle inequality we have $BD < DA + AB = 9 + 5 = 14$, and also $BD + CD > BC$, hence $BD > BC - CD = 17 - 5 = 12$.

We get that $12 < BD < 14$, and as we know that $BD$ is an integer, we must have $BD=\boxed{13}$.

See Also

2009 AMC 12A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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. AMC logo.png