Difference between revisions of "2017 AMC 12A Problems/Problem 24"
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Quadrilateral <math>ABCD</math> is inscribed in circle <math>O</math> and has side lengths <math>AB=3, BC=2, CD=6</math>, and <math>DA=8</math>. Let <math>X</math> and <math>Y</math> be points on <math>\overline{BD}</math> such that <math>\frac{DX}{BD} = \frac{1}{4}</math> and <math>\frac{BY}{BD} = \frac{11}{36}</math>. | Quadrilateral <math>ABCD</math> is inscribed in circle <math>O</math> and has side lengths <math>AB=3, BC=2, CD=6</math>, and <math>DA=8</math>. Let <math>X</math> and <math>Y</math> be points on <math>\overline{BD}</math> such that <math>\frac{DX}{BD} = \frac{1}{4}</math> and <math>\frac{BY}{BD} = \frac{11}{36}</math>. | ||
Let <math>E</math> be the intersection of line <math>AX</math> and the line through <math>Y</math> parallel to <math>\overline{AD}</math>. Let <math>F</math> be the intersection of line <math>CX</math> and the line through <math>E</math> parallel to <math>\overline{AC}</math>. Let <math>G</math> be the point on circle <math>O</math> other than <math>C</math> that lies on line <math>CX</math>. What is <math>XF\cdot XG</math>? | Let <math>E</math> be the intersection of line <math>AX</math> and the line through <math>Y</math> parallel to <math>\overline{AD}</math>. Let <math>F</math> be the intersection of line <math>CX</math> and the line through <math>E</math> parallel to <math>\overline{AC}</math>. Let <math>G</math> be the point on circle <math>O</math> other than <math>C</math> that lies on line <math>CX</math>. What is <math>XF\cdot XG</math>? | ||
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| + | ==See Also== | ||
| + | {{AMC12 box|year=2017|ab=A|num-b=23|num-a=25}} | ||
| + | {{MAA Notice}} | ||
Revision as of 17:24, 8 February 2017
Problem
Quadrilateral 
is inscribed in circle 
and has side lengths 
, and 
. Let 
and 
be points on 
such that 
and 
.
Let 
be the intersection of line 
and the line through parallel to 
. Let 
be the intersection of line 
and the line through parallel to 
. Let 
be the point on circle other than 
that lies on line . What is 
?
See Also
| 2017 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 23 |
Followed by Problem 25 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
