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Difference between revisions of "1993 AHSME Problems/Problem 14"

(Created page with "== Problem == <asy> draw((-1,0)--(1,0)--(1+sqrt(2),sqrt(2))--(0,sqrt(2)+sqrt(13-2*sqrt(2)))--(-1-sqrt(2),sqrt(2))--cycle,black+linewidth(.75)); MP("A",(-1,0),SW);MP("B",(1,0),SE)...")
 
(→‎Solution)
 
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== Solution ==
 
== Solution ==
−
<math>\fbox{B}</math>
+
<asy>
 +
draw((-1,0)--(1,0)--(1+sqrt(2),sqrt(2))--(0,sqrt(2)+sqrt(13-2*sqrt(2)))--(-1-sqrt(2),sqrt(2))--cycle,black+linewidth(.75));
 +
draw((1+sqrt(2),sqrt(2))--(-1-sqrt(2),sqrt(2)));
 +
draw((-1,0)--(-1,sqrt(2)));
 +
draw((1,0)--(1,sqrt(2)));
 +
MP("F",(-1,sqrt(2)),N);MP("G",(1,sqrt(2)),N);
 +
MP("A",(-1,0),SW);MP("B",(1,0),SE);MP("C",(1+sqrt(2),sqrt(2)),E);MP("D",(0,sqrt(2)+sqrt(13-2*sqrt(2))),N);MP("E",(-1-sqrt(2),sqrt(2)),W);
 +
dot((-1,0));dot((1,0));dot((1+sqrt(2),sqrt(2)));dot((-1-sqrt(2),sqrt(2)));dot((0,sqrt(2)+sqrt(13-2*sqrt(2))));
 +
</asy>
 +
 
 +
First, drop perpendiculars from points <math>A</math> and <math>B</math> to segment <math>EC</math>.
 +
 
 +
Since <math>\angle EAB = 120^{\circ}</math>, <math>\angle EAF = 30^{\circ}</math>.
 +
 
 +
This implies that <math>\triangle EAF</math> is a 30-60-90 Triangle, so <math>EF = 1</math> and <math>AF = \sqrt3</math>.
 +
 
 +
Similarly, <math>GB = \sqrt3</math> and <math>GC = 1</math>.
 +
 
 +
Since <math>FABG</math> is a rectangle, <math>FG = AB = 2</math>.
 +
 
 +
Now, notice that since <math>EC = EF + FG + GC = 1+2+1=4</math>, triangle <math>DEC</math> is equilateral.
 +
 
 +
Thus, <math>[ABCDE] = [EABC]+[DCE] = \frac{2+4}{2}(\sqrt3)+\frac{4^2\cdot\sqrt3}{4} = 3\sqrt3+4\sqrt3=\boxed{7\sqrt3 (B)}</math>
 +
 
 +
-AOPS81619
  
 
== See also ==
 
== See also ==

Latest revision as of 18:01, 9 March 2020

Problem

[asy] draw((-1,0)--(1,0)--(1+sqrt(2),sqrt(2))--(0,sqrt(2)+sqrt(13-2*sqrt(2)))--(-1-sqrt(2),sqrt(2))--cycle,black+linewidth(.75)); MP("A",(-1,0),SW);MP("B",(1,0),SE);MP("C",(1+sqrt(2),sqrt(2)),E);MP("D",(0,sqrt(2)+sqrt(13-2*sqrt(2))),N);MP("E",(-1-sqrt(2),sqrt(2)),W); dot((-1,0));dot((1,0));dot((1+sqrt(2),sqrt(2)));dot((-1-sqrt(2),sqrt(2)));dot((0,sqrt(2)+sqrt(13-2*sqrt(2)))); [/asy]

The convex pentagon $ABCDE$ has $\angle{A}=\angle{B}=120^\circ,EA=AB=BC=2$ and $CD=DE=4$. What is the area of ABCDE?

$\text{(A) } 10\quad \text{(B) } 7\sqrt{3}\quad \text{(C) } 15\quad \text{(D) } 9\sqrt{3}\quad \text{(E) } 12\sqrt{5}$

Solution

[asy] draw((-1,0)--(1,0)--(1+sqrt(2),sqrt(2))--(0,sqrt(2)+sqrt(13-2*sqrt(2)))--(-1-sqrt(2),sqrt(2))--cycle,black+linewidth(.75)); draw((1+sqrt(2),sqrt(2))--(-1-sqrt(2),sqrt(2))); draw((-1,0)--(-1,sqrt(2))); draw((1,0)--(1,sqrt(2))); MP("F",(-1,sqrt(2)),N);MP("G",(1,sqrt(2)),N); MP("A",(-1,0),SW);MP("B",(1,0),SE);MP("C",(1+sqrt(2),sqrt(2)),E);MP("D",(0,sqrt(2)+sqrt(13-2*sqrt(2))),N);MP("E",(-1-sqrt(2),sqrt(2)),W); dot((-1,0));dot((1,0));dot((1+sqrt(2),sqrt(2)));dot((-1-sqrt(2),sqrt(2)));dot((0,sqrt(2)+sqrt(13-2*sqrt(2)))); [/asy]

First, drop perpendiculars from points $A$ and $B$ to segment $EC$.

Since $\angle EAB = 120^{\circ}$, $\angle EAF = 30^{\circ}$.

This implies that $\triangle EAF$ is a 30-60-90 Triangle, so $EF = 1$ and $AF = \sqrt3$.

Similarly, $GB = \sqrt3$ and $GC = 1$.

Since $FABG$ is a rectangle, $FG = AB = 2$.

Now, notice that since $EC = EF + FG + GC = 1+2+1=4$, triangle $DEC$ is equilateral.

Thus, $[ABCDE] = [EABC]+[DCE] = \frac{2+4}{2}(\sqrt3)+\frac{4^2\cdot\sqrt3}{4} = 3\sqrt3+4\sqrt3=\boxed{7\sqrt3 (B)}$

-AOPS81619

See also

1993 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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
All AHSME Problems and Solutions

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