Difference between revisions of "1965 AHSME Problems/Problem 16"

Latest revision as of 15:58, 18 July 2024

Problem

Let line $AC$ be perpendicular to line $CE$ . Connect $A$ to $D$ , the midpoint of $CE$ , and connect $E$ to $B$ , the midpoint of $AC$ . If $AD$ and $EB$ intersect in point $F$ , and $\overline{BC} = \overline{CD} = 15$ inches, then the area of triangle $DFE$ , in square inches, is:

$\textbf{(A)}\ 50 \qquad \textbf{(B) }\ 50\sqrt {2} \qquad \textbf{(C) }\ 75 \qquad \textbf{(D) }\ \frac{15}{2}\sqrt{105}\qquad \textbf{(E) }\ 100$

Solution

$[asy] draw((0,0)--(30,0)); dot((0,0)); label("C", (-2,-2)); dot((30,0)); label("E", (32,-2)); dot((15,0)); label("D",(15,-2)); dot((0,15)); label("B",(-2,15)); draw((0,0)--(0,30)--(30,0)); dot((0,30)); label("A",(-2, 32)); dot((15,15)); label("G", (17,17)); draw((0,0)--(15,15)); dot((10,10)); label("F", (10,12)); draw((0,15)--(30,0)); draw((15,0)--(0,30)); markscalefactor=0.25; draw(rightanglemark((0,30),(0,0),(30,0))); label("$15$",(-2,7.5)); label("$15$",(-2,22.5)); label("$15$",(7.5,-2)); label("$15$",(22.5,-2)); [/asy]$

Draw $\overline{AE}$ , as seen in the diagram. From the problem, we know that $\overline{EB}$ and $\overline{AD}$ are medians of $\triangle ACE$ . Let $G$ be the midpoint of $\overline{AE}$ . Then, $\overline{CG}$ is also a median of $\triangle ACE$ , and it goes through $\triangle ACE$ 's centroid, $F$ . Because medians divide their triangle into $6$ smaller triangles of equal area, we know that $[\triangle DFE]=\frac{1}{6}[\triangle ACE]$ . Because $[\triangle ACE]=\frac{1}{2}*(15+15)^2=\frac{900}{2}=450$ , $[\triangle DFE]=\frac{450}{6}=75$ . Thus, our answer is $\boxed{\textbf{(C) }75}$ .

1965 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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Difference between revisions of "1965 AHSME Problems/Problem 16"

Latest revision as of 15:58, 18 July 2024

Problem

Solution

See Also

@@ Line 12: / Line 12: @@
 == Solution ==
-<math>\fbox{C}</math>
+<asy>
+draw((0,0)--(30,0));
+dot((0,0));
+label("C", (-2,-2));
+dot((30,0));
+label("E", (32,-2));
+dot((15,0));
+label("D",(15,-2));
+dot((0,15));
+label("B",(-2,15));
+draw((0,0)--(0,30)--(30,0));
+dot((0,30));
+label("A",(-2, 32));
+dot((15,15));
+label("G", (17,17));
+draw((0,0)--(15,15));
+dot((10,10));
+label("F", (10,12));
+draw((0,15)--(30,0));
+draw((15,0)--(0,30));
+markscalefactor=0.25;
+draw(rightanglemark((0,30),(0,0),(30,0)));
+label("$15$",(-2,7.5));
+label("$15$",(-2,22.5));
+label("$15$",(7.5,-2));
+label("$15$",(22.5,-2));
+</asy>
+Draw <math>\overline{AE}</math>, as seen in the diagram. From the problem, we know that <math>\overline{EB}</math> and <math>\overline{AD}</math> are [[median of a triangle|medians]] of <math>\triangle ACE</math>. Let <math>G</math> be the midpoint of <math>\overline{AE}</math>. Then, <math>\overline{CG}</math> is also a median of <math>\triangle ACE</math>, and it goes through <math>\triangle ACE</math>'s [[centroid]], <math>F</math>. Because medians divide their triangle into <math>6</math> smaller triangles of equal area, we know that <math>[\triangle DFE]=\frac{1}{6}[\triangle ACE]</math>. Because <math>[\triangle ACE]=\frac{1}{2}*(15+15)^2=\frac{900}{2}=450</math>, <math>[\triangle DFE]=\frac{450}{6}=75</math>. Thus, our answer is <math>\boxed{\textbf{(C) }75}</math>.
 ==See Also==
 {{AHSME 40p box|year=1965|num-b=15|num-a=17}}
+{{MAA Notice}}
 [[Category:Introductory Geometry Problems]]