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 "1959 AHSME Problems/Problem 40"

(Add problem statement & Unify answer)
m (→‎Solution 2: fixed type; changed answer to second solution to 15, same letter.)
 
(6 intermediate revisions by one other user not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
  
−
On the same side of a straight line three circles are drawn as follows: a circle with a radius of <math>4</math> inches is tangent to the line, the other two circles are equal, and each is tangent to the line and to the other two circles. The radius of the equal circles is: <math>\textbf{(A)}\ 24 \qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 12</math>
+
In <math>\triangle ABC</math>, <math>BD</math> is a median. <math>CF</math> intersects <math>BD</math> at <math>E</math> so that <math>\overline{BE}=\overline{ED}</math>. Point <math>F</math> is on <math>AB</math>. Then, if <math>\overline{BF}=5</math>,  
 +
<math>\overline{BA}</math> equals:
 +
<math>\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ \text{none of these}  </math>
  
−
== Solution ==
+
== Solution 1 ==
  
−
Make a line DG which is parallel to FC. We know that GF = BF = 5, since BFE is similar to BGD.
+
<asy>
  
−
Since ADG is similar to ACB, we know that AG is 10.
+
import geometry;
  
−
Thus, AF = 10 + 5 = 15, which is answer <math>\boxed{C}</math>
+
point A = (0,0);
 +
point B = (5,8);
 +
point C = (16,0);
 +
point D = midpoint(A--C);
 +
point E = midpoint(B--D);
 +
point F, G;
 +
triangle ABC = triangle(A,B,C);
 +
 
 +
// Triangle ABC
 +
draw(ABC);
 +
dot(A);
 +
label("A",A,SW);
 +
dot(B);
 +
label("B",B,N);
 +
dot(C);
 +
label("C",C,SE);
 +
 
 +
// Segment BD
 +
draw(B--D);
 +
dot(D);
 +
label("D",D,S);
 +
dot(E);
 +
label("E",E,SW);
 +
 
 +
// Segment CF
 +
pair[] f = intersectionpoints(line(C,E),A--B);
 +
F = f[0];
 +
dot(F);
 +
label("F",F,W);
 +
draw(C--F);
 +
 
 +
// Segment DG
 +
pair[] g = intersectionpoints(parallel(D,line(F,C)),A--B);
 +
G = g[0];
 +
dot(G);
 +
label("G",G,W);
 +
draw(D--G);
 +
 
 +
// Length Label
 +
label("$5$", midpoint(B--F), NW);
 +
 
 +
</asy>
 +
 
 +
Draw <math>\overline{DG} \parallel \overline{FC}</math> with <math>G</math> on <math>\overline{AB}</math>. We know that <math>GF = BF = 5</math>, since <math>\triangle BFE \sim \triangle BGD</math>.
 +
 
 +
Likewise, since <math>\triangle ADG \sim \triangle ACF</math>, we know that <math>AG=5</math>.
 +
 
 +
Thus, <math>AF=AG+GF+FB=5+5+5=15</math>, which is answer <math>\fbox{\textbf{(C)}}</math>.
 +
 
 +
 
 +
== Solution 2 ==
 +
 
 +
Let <math>AD=DC=x</math> and <math>BE=ED=y</math>. By [[Menelaus' Theorem]] on <math>\triangle BEF</math> and <math>\overleftrightarrow{AC}</math>, we know the following:
 +
\begin{align*}
 +
\frac{BD}{ED}*\frac{EC}{FC}*\frac{FA}{BA} &= 1 \\
 +
\frac{2y}{y}*\frac{EC}{FC}*\frac{FA}{5+FA} &= 1 \\
 +
\frac{EC}{FC}*\frac{FA}{5+FA} &= \frac{1}{2}
 +
\end{align*}
 +
By applying Menelaus again on <math>\triangle CDE</math> and <math>\overleftrightarrow{AB}</math>, we see that:
 +
\begin{align*}
 +
\frac{CA}{DA}*\frac{DB}{EB}*\frac{EF}{CF} &= 1 \\
 +
\frac{2x}{x}*\frac{2y}{y}*\frac{CF-EC}{FC} &= 1 \\
 +
1-\frac{EC}{FC} &= \frac{1}{4} \\
 +
\frac{EC}{FC} &= \frac{3}{4}
 +
\end{align*}
 +
Substituting <math>\frac{3}{4}</math> for <math>\frac{EC}{FC}</math> into the previous equation, we can now solve for <math>FA</math>:
 +
\begin{align*}
 +
\frac{EC}{FC}*\frac{FA}{5+FA} &= \frac{1}{2} \\
 +
\frac{3}{4}*\frac{FA}{5+FA} &= \frac{1}{2} \\
 +
\frac{FA}{5+FA} &= \frac{2}{3} \\
 +
3FA &= 10+2FA \\
 +
FA &= 10
 +
\end{align*}
 +
Because <math>AB=AF+FB</math>, <math>AB=5+10=\boxed{\textbf{(C) } 15}</math>.
 +
 
 +
== See also ==
 +
{{AHSME 50p box|year=1959|num-b=39|num-a=41}}
 +
{{MAA Notice}}
 +
[[Category:Intermediate Geometry Problems]]

Latest revision as of 08:39, 16 October 2024

Problem

In $\triangle ABC$, $BD$ is a median. $CF$ intersects $BD$ at $E$ so that $\overline{BE}=\overline{ED}$. Point $F$ is on $AB$. Then, if $\overline{BF}=5$, $\overline{BA}$ equals: $\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ \text{none of these}$

Solution 1

[asy] import geometry; point A = (0,0); point B = (5,8); point C = (16,0); point D = midpoint(A--C); point E = midpoint(B--D); point F, G; triangle ABC = triangle(A,B,C); // Triangle ABC draw(ABC); dot(A); label("A",A,SW); dot(B); label("B",B,N); dot(C); label("C",C,SE); // Segment BD draw(B--D); dot(D); label("D",D,S); dot(E); label("E",E,SW); // Segment CF pair[] f = intersectionpoints(line(C,E),A--B); F = f[0]; dot(F); label("F",F,W); draw(C--F); // Segment DG pair[] g = intersectionpoints(parallel(D,line(F,C)),A--B); G = g[0]; dot(G); label("G",G,W); draw(D--G); // Length Label label("$5$", midpoint(B--F), NW); [/asy]

Draw $\overline{DG} \parallel \overline{FC}$ with $G$ on $\overline{AB}$. We know that $GF = BF = 5$, since $\triangle BFE \sim \triangle BGD$.

Likewise, since $\triangle ADG \sim \triangle ACF$, we know that $AG=5$.

Thus, $AF=AG+GF+FB=5+5+5=15$, which is answer $\fbox{\textbf{(C)}}$.


Solution 2

Let $AD=DC=x$ and $BE=ED=y$. By Menelaus' Theorem on $\triangle BEF$ and $\overleftrightarrow{AC}$, we know the following: \begin{align*} \frac{BD}{ED}*\frac{EC}{FC}*\frac{FA}{BA} &= 1 \\ \frac{2y}{y}*\frac{EC}{FC}*\frac{FA}{5+FA} &= 1 \\ \frac{EC}{FC}*\frac{FA}{5+FA} &= \frac{1}{2} \end{align*} By applying Menelaus again on $\triangle CDE$ and $\overleftrightarrow{AB}$, we see that: \begin{align*} \frac{CA}{DA}*\frac{DB}{EB}*\frac{EF}{CF} &= 1 \\ \frac{2x}{x}*\frac{2y}{y}*\frac{CF-EC}{FC} &= 1 \\ 1-\frac{EC}{FC} &= \frac{1}{4} \\ \frac{EC}{FC} &= \frac{3}{4} \end{align*} Substituting $\frac{3}{4}$ for $\frac{EC}{FC}$ into the previous equation, we can now solve for $FA$: \begin{align*} \frac{EC}{FC}*\frac{FA}{5+FA} &= \frac{1}{2} \\ \frac{3}{4}*\frac{FA}{5+FA} &= \frac{1}{2} \\ \frac{FA}{5+FA} &= \frac{2}{3} \\ 3FA &= 10+2FA \\ FA &= 10 \end{align*} Because $AB=AF+FB$, $AB=5+10=\boxed{\textbf{(C) } 15}$.

See also

1959 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 39 		Followed by
Problem 41
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=1959_AHSME_Problems/Problem_40&oldid=230003"