Difference between revisions of "2016 AMC 10B Problems/Problem 19"

Revision as of 09:29, 21 February 2016

Problem

Rectangle $ABCD$ has $AB=5$ and $BC=4$ . Point $E$ lies on $\overline{AB}$ so that $EB=1$ , point $G$ lies on $\overline{BC}$ so that $CG=1$ . and point $F$ lies on $\overline{CD}$ so that $DF=2$ . Segments $\overline{AG}$ and $\overline{AC}$ intersect $\overline{EF}$ at $Q$ and $P$ , respectively. What is the value of $\frac{PQ}{EF}$ ?

$[asy]pair A1=(2,0),A2=(4,4); pair B1=(0,4),B2=(5,1); pair C1=(5,0),C2=(0,4); draw(A1--A2); draw(B1--B2); draw(C1--C2); draw((0,0)--B1--(5,4)--C1--cycle); dot((20/7,12/7)); dot((3.07692307692,2.15384615384)); label("$Q$",(3.07692307692,2.15384615384),N); label("$P$",(20/7,12/7),W); label("$A$",(0,4), NW); label("$B$",(5,4), NE); label("$C$",(5,0),SE); label("$D$",(0,0),SW); label("$F$",(2,0),S); label("$G$",(5,1),E); label("$E$",(4,4),N);[/asy]$

$\textbf{(A)}~\frac{\sqrt{13}}{16} \qquad \textbf{(B)}~\frac{\sqrt{2}}{13} \qquad \textbf{(C)}~\frac{9}{82} \qquad \textbf{(D)}~\frac{10}{91}\qquad \textbf{(E)}~\frac19$

Solution

$\textbf{(D)}~\frac{10}{91}$

solution by ngeorge

@@ Line 26: / Line 26: @@
 \textbf{(D)}~\frac{10}{91}\qquad
 \textbf{(E)}~\frac19</math>
+==Solution==
+<math>\textbf{(D)}~\frac{10}{91}</math>
+solution by ngeorge

Page

Toolbox

Search

Difference between revisions of "2016 AMC 10B Problems/Problem 19"

Revision as of 09:29, 21 February 2016

Problem

Solution