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Difference between revisions of "1959 AHSME Problems/Problem 40"

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<math>\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ \text{none of these}  </math>
 
<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.
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<asy>
  
−
Since ADG is similar to ACB, we know that AG is 10.
+
import geometry;
  
−
Thus, AF = 10 + 5 = 15, which is answer <math>\fbox{\textbf{(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];
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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) } 10}</math>.
  
 
== See also ==
 
== See also ==
 
{{AHSME 50p box|year=1959|num-b=39|num-a=41}}
 
{{AHSME 50p box|year=1959|num-b=39|num-a=41}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
[[Category:Intermediate Geometry Problems]]

Revision as of 18:05, 21 July 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 $AB=AF+FB$, $AB=5+10=\boxed{\textbf{(C) } 10}$.

See also

1959 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 39 		Followed by
Problem 41
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