Difference between revisions of "1986 AHSME Problems/Problem 16"

(Created page with "==Problem== In <math>\triangle ABC, AB = 8, BC = 7, CA = 6</math> and side <math>BC</math> is extended, as shown in the figure, to a point <math>P</math> so that <math>\triangl...")
 
m (See also)
Line 30: Line 30:
 
{{AHSME box|year=1986|num-b=15|num-a=17}}   
 
{{AHSME box|year=1986|num-b=15|num-a=17}}   
  
[[Category: Inttermediate Geometry Problems]]
+
[[Category: Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 02:37, 24 October 2014

Problem

In $\triangle ABC, AB = 8, BC = 7, CA = 6$ and side $BC$ is extended, as shown in the figure, to a point $P$ so that $\triangle PAB$ is similar to $\triangle PCA$. The length of $PC$ is

[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, P=(1.5,5), B=(8,0), C=P+2.5*dir(P--B); draw(A--P--C--A--B--C); label("A", A, W); label("B", B, E); label("C", C, NE); label("P", P, NW); label("6", 3*dir(A--C), SE); label("7", B+3*dir(B--C), NE); label("8", (4,0), S); [/asy]

$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 8\qquad \textbf{(C)}\ 9\qquad \textbf{(D)}\ 10\qquad \textbf{(E)}\ 11$

Solution

See also

1986 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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 26 27 28 29 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png