Difference between revisions of "1959 AHSME Problems/Problem 40"
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== Problem == | == Problem == | ||
− | + | 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 == |
− | + | <asy> | |
− | + | import geometry; | |
− | Thus, AF = | + | 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) } 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]] |
Latest revision as of 19:05, 21 July 2024
Contents
Problem
In ,
is a median.
intersects
at
so that
. Point
is on
. Then, if
,
equals:
Solution 1
Draw with
on
. We know that
, since
.
Likewise, since , we know that
.
Thus, , which is answer
.
Solution 2
Let and
. By Menelaus' Theorem on
and
, 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
and
, 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
for
into the previous equation, we can now solve for
:
\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
,
.
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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