Difference between revisions of "1965 AHSME Problems/Problem 37"

Latest revision as of 09:38, 20 July 2024

Problem

Point $E$ is selected on side $AB$ of $\triangle{ABC}$ in such a way that $AE: EB = 1: 3$ and point $D$ is selected on side $BC$ such that $CD: DB = 1: 2$ . The point of intersection of $AD$ and $CE$ is $F$ . Then $\frac {EF}{FC} + \frac {AF}{FD}$ is:

$\textbf{(A)}\ \frac {4}{5} \qquad \textbf{(B) }\ \frac {5}{4} \qquad \textbf{(C) }\ \frac {3}{2} \qquad \textbf{(D) }\ 2\qquad \textbf{(E) }\ \frac{5}{2}$

Solution

$[asy] import geometry; point A = (0,0); point B = (16,0); point C = (3, 10); point D, E, F; real d; // Triangle ABC draw(A--B--C--A); dot(A); label("A", A, SW); dot(B); label("B", B, SE); dot(C); label("C", C, NW); // Segments AD and CE D = 2/3*C+1/3*B; dot(D); label("D", D, NE); draw(A--D); E = midpoint(A--midpoint(A--B)); dot(E); label("E", E, S); draw(C--E); // Point F pair[] f=intersectionpoints((A--D), (C--E)); F=f[0]; dot(F); label("F", F, SE); [/asy]$

We use mass points for this problem. Let $\text{m} A$ denote the mass of point $A$ . Rewrite the expression we are finding as $\frac{EF}{FC} + \frac{AF}{FD} = \frac{FE}{FC} + \frac{FA}{FD} = \frac{\text{m} C}{\text{m} E} + \frac{\text{m} D}{\text{m} A}$ Now, let $\text{m} C = 2$ . We then have $2 \cdot 1 = \text{m} B \cdot 2$ , so $\text{m} B = 1$ , and $\text{m} D = 2+1 = 3$ We can let $\text{m} A = 3$ . We have $\text{m} E = \text{m} A + \text{m} B = 3+1 = 4$ From here, substitute the respective values to get

$\frac{\text{m} C}{\text{m} E} + \frac{\text{m} D}{\text{m} A} = \frac{2}{4} + \frac{3}{3} = \frac{1}{2} + 1 = \frac{3}{2}$ This answer corresponds to $\fbox{\textbf{(C)}}$ .

~JustinLee2017

1965 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 36	Followed by Problem 38
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Difference between revisions of "1965 AHSME Problems/Problem 37"

Latest revision as of 09:38, 20 July 2024

Problem

Solution

See Also

@@ Line 11: / Line 11: @@
 == Solution ==
+<asy>
+import geometry;
+point A = (0,0);
+point B = (16,0);
+point C = (3, 10);
+point D, E, F;
+real d;
+// Triangle ABC
+draw(A--B--C--A);
+dot(A);
+label("A", A, SW);
+dot(B);
+label("B", B, SE);
+dot(C);
+label("C", C, NW);
+// Segments AD and CE
+D = 2/3*C+1/3*B;
+dot(D);
+label("D", D, NE);
+draw(A--D);
+E = midpoint(A--midpoint(A--B));
+dot(E);
+label("E", E, S);
+draw(C--E);
+// Point F
+pair[] f=intersectionpoints((A--D), (C--E));
+F=f[0];
+dot(F);
+label("F", F, SE);
+</asy>
 We use [[mass points]] for this problem. Let <math>\text{m} A</math> denote the mass of point <math>A</math>.
@@ Line 20: / Line 57: @@
 <cmath>\frac{\text{m} C}{\text{m} E} + \frac{\text{m} D}{\text{m} A} = \frac{2}{4} + \frac{3}{3} = \frac{1}{2} + 1 = \frac{3}{2}</cmath>
-This answer corresponds to answer <math>\fbox{\textbf{(C)}}</math>.
+This answer corresponds to <math>\fbox{\textbf{(C)}}</math>.
 ~JustinLee2017