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

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m (→‎Solution 2: fixed type; changed answer to second solution to 15, same letter.)
 
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FA &= 10
 
FA &= 10
 
\end{align*}
 
\end{align*}
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Because <math>AB=AF+FB</math>, <math>AB=5+10=\boxed{\textbf{(C) } 10}</math>.
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Because <math>AB=AF+FB</math>, <math>AB=5+10=\boxed{\textbf{(C) } 15}</math>.
  
 
== See also ==
 
== See also ==

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

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