Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1952 AHSME Problems/Problem 29"

(→‎Solution)
(→‎Solution)
 
Line 10: Line 10:
  
 
== Solution ==
 
== Solution ==
−
<math>\fbox{}</math>
+
 
  
  
Line 19: Line 19:
 
It follows that <math>a = \sqrt{x^2 - 25} = 5\sqrt{\frac{25}{16}-1} = \frac{15}{4}</math>.
 
It follows that <math>a = \sqrt{x^2 - 25} = 5\sqrt{\frac{25}{16}-1} = \frac{15}{4}</math>.
  
−
Thus, the answer is <math>E</math>.
+
<math>BN = 5 + \frac{15}{14} = 8.75</math>, so the answer is <math>A</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 13:20, 15 January 2018

Problem

In a circle of radius $5$ units, $CD$ and $AB$ are perpendicular diameters. A chord $CH$ cutting $AB$ at $K$ is $8$ units long. The diameter $AB$ is divided into two segments whose dimensions are:

$\textbf{(A)}\ 1.25, 8.75 \qquad \textbf{(B)}\ 2.75,7.25 \qquad \textbf{(C)}\ 2,8 \qquad \textbf{(D)}\ 4,6 \qquad \textbf{(E)}\ \text{none of these}$

Solution

Let the intersection of $CH$ and $AB$ be $N$, $O$ be the center of the circle, $ON = a$ and $CN = x$. By power of a point on $N$, we have \[(5+a)(5-a) = 25-a^2 = x(8-x).\] $\triangle CON$ is a right triangle, so we also know that $x^2 = a^2 + 25$, thus \[25 - a^2 = 25-(x^2-25) = 50-x^2 = 8x-x^2 \Rightarrow x = \frac{50}{8} = \frac{25}{4}.\] It follows that $a = \sqrt{x^2 - 25} = 5\sqrt{\frac{25}{16}-1} = \frac{15}{4}$.

$BN = 5 + \frac{15}{14} = 8.75$, so the answer is $A$.

See also

1952 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 28 		Followed by
Problem 30
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1952_AHSME_Problems/Problem_29&oldid=89784"