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

Revision as of 20:58, 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 $\boxed{014}$ .

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

Page

Toolbox

Search

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

Revision as of 20:58, 12 July 2009

Problem

Solution

See also

@@ Line 4: / Line 4: @@
 == Solution ==
 <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));
+pair B = (0,0), C = (25,0), A = (578/50,19.8838);
-D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle);
+draw(A--B--C--cycle);
-pair D=(B+C)/2,E=(A+C)/2,F=(A+B)/2;
+label("$B$",B,SW); label("$C$",C,SE); label("$A$",A,N);
-D(circumcircle(MP("D",D),MP("E",E),C));
+pair D = (13,0), E = (11*A + 13*C)/24, F = (12*B + 11*A)/23;
-D(circumcircle(B,MP("F",F),D));
+draw(D--E--F--cycle);
-D(circumcircle(A,F,E));
+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></center>
-{{solution}}
+From adjacent sides, the following relationships can be derived:
+<math>\begin{align*}
+DC &= EC + 1\\
+AE &= AF + 1\\
+BD &= BF + 2
+\end{align*}</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>.
 == See also ==