Difference between revisions of "Mock AIME 1 2007-2008 Problems/Problem 11"
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From adjacent sides, the following relationships can be derived: | From adjacent sides, the following relationships can be derived: | ||
| − | < | + | <cmath>\begin{align*} |
DC &= EC + 1\\ | DC &= EC + 1\\ | ||
AE &= AF + 1\\ | AE &= AF + 1\\ | ||
BD &= BF + 2 | BD &= BF + 2 | ||
| − | \end{align*}</ | + | \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 20:02, 29 January 2020
Problem
is inscribed inside 
such that 
lie on 
, respectively. The circumcircles of 
have centers 
, respectively. Also, 
, and 
. The length of 
can be written in the form 
, where 
and 
are relatively prime integers. Find 
.
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]](http://latex.artofproblemsolving.com/6/3/e/63e4e9b2ac599b49eec642e212fd1dafc986a5e4.png)
From adjacent sides, the following relationships can be derived:
Since 
, and 
, 
. Thus, 
. 
. Thus, 
. Thus, the answer is 
.
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 | ||
