Difference between revisions of "Mock AIME 1 2007-2008 Problems/Problem 11"

(Solution: start asy)
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== Solution ==
 
== Solution ==
 
<center><asy>
 
<center><asy>
pointpen = black; pathpen = black;
+
size(150); defaultpen(linewidth(0.8)); import markers; import geometry_dev;
pair A=(0,0),B=(23,0),C=IP(Circle(A,24),Circle(B,25));
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pair B = (0,0), C = (25,0), A = (578/50,19.8838);
D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle);
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draw(A--B--C--cycle);
pair D=(B+C)/2,E=(A+C)/2,F=(A+B)/2;
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label("$B$",B,SW); label("$C$",C,SE); label("$A$",A,N);
D(circumcircle(MP("D",D),MP("E",E),C));
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pair D = (13,0), E = (11*A + 13*C)/24, F = (12*B + 11*A)/23;
D(circumcircle(B,MP("F",F),D));
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draw(D--E--F--cycle);
D(circumcircle(A,F,E));
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label("$D$",D,dir(-90));
 +
label("$E$",E,dir(0));
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label("$F$",F,dir(180));
 +
 
 +
draw(A--E,StickIntervalMarker(1,3,size=6));draw(B--D,StickIntervalMarker(1,3,size=6));
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draw(F--B,StickIntervalMarker(1,2,size=6)); draw(E--C,StickIntervalMarker(1,2,size=6));
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draw(A--F,StickIntervalMarker(1,1,size=6)); draw(C--D,StickIntervalMarker(1,1,size=6));
 +
 
 +
label("24",A--C,5*dir(0)); label("25",B--C,5*dir(-90)); label("23",B--A,5*dir(180));
 
</asy></center>
 
</asy></center>
{{solution}}
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 +
From adjacent sides, the following relationships can be derived:
 +
 
 +
<math>\begin{align*}
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DC &= EC + 1\\
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AE &= AF + 1\\
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BD &= BF + 2
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\end{align*}</math>
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 +
Since <math>BF = EC</math>, and <math>DC = BF + 1</math>, <math>BD = DC + 1</math>. Thus, <math>BC = BD + DC = BD + (BD - 1)</math>. <math>26 = 2BD</math>. Thus, <math>BD = 13/1</math>. Thus, the answer is <math>\boxed{014}</math>.
 +
 
  
 
== See also ==
 
== See also ==

Revision as of 21:58, 12 July 2009

Problem

$\triangle DEF$ is inscribed inside $\triangle ABC$ such that $D,E,F$ lie on $BC, AC, AB$, respectively. The circumcircles of $\triangle DEC, \triangle BFD, \triangle AFE$ have centers $O_1,O_2,O_3$, respectively. Also, $AB = 23, BC = 25, AC=24$, and $\stackrel{\frown}{BF} = \stackrel{\frown}{EC},\ \stackrel{\frown}{AF} = \stackrel{\frown}{CD},\ \stackrel{\frown}{AE} = \stackrel{\frown}{BD}$. The length of $BD$ can be written in the form $\frac mn$, where $m$ and $n$ are relatively prime integers. Find $m+n$.

Solution

size(150); defaultpen(linewidth(0.8)); import markers; import geometry_dev;
pair B = (0,0), C = (25,0), A = (578/50,19.8838);
draw(A--B--C--cycle);
label("$B$",B,SW); label("$C$",C,SE); label("$A$",A,N);
pair D = (13,0), E = (11*A + 13*C)/24, F = (12*B + 11*A)/23;
draw(D--E--F--cycle);
label("$D$",D,dir(-90));
label("$E$",E,dir(0));
label("$F$",F,dir(180));

draw(A--E,StickIntervalMarker(1,3,size=6));draw(B--D,StickIntervalMarker(1,3,size=6));
draw(F--B,StickIntervalMarker(1,2,size=6)); draw(E--C,StickIntervalMarker(1,2,size=6));
draw(A--F,StickIntervalMarker(1,1,size=6)); draw(C--D,StickIntervalMarker(1,1,size=6));

label("24",A--C,5*dir(0)); label("25",B--C,5*dir(-90)); label("23",B--A,5*dir(180));
 (Error making remote request. Unknown error_msg)

From adjacent sides, the following relationships can be derived:

$\begin{align*} DC &= EC + 1\\ AE &= AF + 1\\ BD &= BF + 2 \end{align*}$ (Error compiling LaTeX. Unknown error_msg)

Since $BF = EC$, and $DC = BF + 1$, $BD = DC + 1$. Thus, $BC = BD + DC = BD + (BD - 1)$. $26 = 2BD$. Thus, $BD = 13/1$. Thus, the answer is $\boxed{014}$.


See also

Mock AIME 1 2007-2008 (Problems, Source)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15