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

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== Solution ==
 
== Solution ==
−
<center></asy>
+
<center><asy>
−
size(150); defaultpen(linewidth(0.8)); import markers; import geometry_dev;
+
size(150); defaultpen(linewidth(0.8)); import markers;
 
pair B = (0,0), C = (25,0), A = (578/50,19.8838);
 
pair B = (0,0), C = (25,0), A = (578/50,19.8838);
 
draw(A--B--C--cycle);
 
draw(A--B--C--cycle);
−
label("<math>B</math>",B,SW); label("<math>C</math>",C,SE); label("<math>A</math>",A,N);
+
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;
 
pair D = (13,0), E = (11*A + 13*C)/24, F = (12*B + 11*A)/23;
 
draw(D--E--F--cycle);
 
draw(D--E--F--cycle);
−
label("<math>D</math>",D,dir(-90));
+
label("$D$",D,dir(-90));
−
label("<math>E</math>",E,dir(0));
+
label("$E$",E,dir(0));
−
label("<math>F</math>",F,dir(180));
+
label("$F$",F,dir(180));
  
 
draw(A--E,StickIntervalMarker(1,3,size=6));draw(B--D,StickIntervalMarker(1,3,size=6));
 
draw(A--E,StickIntervalMarker(1,3,size=6));draw(B--D,StickIntervalMarker(1,3,size=6));
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From adjacent sides, the following relationships can be derived:
 
From adjacent sides, the following relationships can be derived:
  
−
<math>\begin{align*}
+
<cmath>\begin{align*}
 
DC &= EC + 1\\
 
DC &= EC + 1\\
 
AE &= AF + 1\\
 
AE &= AF + 1\\
 
BD &= BF + 2
 
BD &= BF + 2
−
\end{align*}</math>
+
\end{align*}</cmath>
  
 
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>.
 
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>.

Latest revision as of 19:02, 29 January 2020

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

[asy] size(150); defaultpen(linewidth(0.8)); import markers; 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)); [/asy]

From adjacent sides, the following relationships can be derived:

\begin{align*} DC &= EC + 1\\ AE &= AF + 1\\ BD &= BF + 2 \end{align*}

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
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