Difference between revisions of "2002 AMC 10B Problems/Problem 17"
(New page: 17. A regular octagon <math>ABCDEFGH</math> has sides of length two. Find the area of <math>\triangle ADG</math>. <math>\textbf{(A) } 4 + 2\sqrt2 \qquad \textbf{(B) } 6 + \sqrt2\qquad \t...) |
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| − | + | == Problem == | |
| + | |||
| + | A regular octagon <math>ABCDEFGH</math> has sides of length two. Find the area of <math>\triangle ADG</math>. | ||
<math>\textbf{(A) } 4 + 2\sqrt2 \qquad \textbf{(B) } 6 + \sqrt2\qquad \textbf{(C) } 4 + 3\sqrt2 \qquad \textbf{(D) } 3 + 4\sqrt2 \qquad \textbf{(E) } 8 + \sqrt2</math> | <math>\textbf{(A) } 4 + 2\sqrt2 \qquad \textbf{(B) } 6 + \sqrt2\qquad \textbf{(C) } 4 + 3\sqrt2 \qquad \textbf{(D) } 3 + 4\sqrt2 \qquad \textbf{(E) } 8 + \sqrt2</math> | ||
| + | |||
| + | == Solution == | ||
| + | |||
| + | <asy> | ||
| + | unitsize(1cm); | ||
| + | defaultpen(0.8); | ||
| + | pair[] A = new pair[8]; | ||
| + | A[0]=(0,0); | ||
| + | for (int i=1; i<8; ++i) A[i] = A[i-1] + 2*dir(45*(i-1)); | ||
| + | draw( A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle ); | ||
| + | label("$A$",A[0],SW); | ||
| + | label("$B$",A[1],SE); | ||
| + | label("$C$",A[2],SE); | ||
| + | label("$D$",A[3],NE); | ||
| + | label("$E$",A[4],NE); | ||
| + | label("$F$",A[5],NW); | ||
| + | label("$G$",A[6],NW); | ||
| + | label("$H$",A[7],SW); | ||
| + | filldraw( A[0]--A[3]--A[6]--cycle, lightgray, black ); | ||
| + | pair P = intersectionpoint( A[3]--A[6], A[0]--A[5] ); | ||
| + | draw( A[0]--P ); | ||
| + | draw( P -- A[5], dashed ); | ||
| + | label("$P$",P,NE); | ||
| + | draw( A[1]--A[4], dashed ); | ||
| + | pair Q = intersectionpoint( A[3]--A[6], A[1]--A[4] ); | ||
| + | label("$Q$",Q,NW); | ||
| + | </asy> | ||
| + | |||
| + | The area of the triangle <math>ADG</math> can be computed as <math>\frac{DG \cdot AP}2</math>. We will now find <math>DG</math> and <math>AP</math>. | ||
| + | |||
| + | Clearly, <math>PFG</math> is a right isosceles triangle with hypotenuse of length <math>2</math>, hence <math>PG=\sqrt 2</math>. | ||
| + | The same holds for triangle <math>QED</math> and its leg <math>QD</math>. The length of <math>PQ</math> is equal to <math>FE=2</math>. | ||
| + | Hence <math>GD = 2 + 2\sqrt 2</math>, and <math>AP = PD = 2 + \sqrt 2</math>. | ||
| + | |||
| + | Then the area of <math>ADG</math> equals <math>\frac{DG \cdot AP}2 = \frac{(2+2\sqrt 2)(2+\sqrt 2)}2 = \frac{8+6\sqrt 2}2 = \boxed{\textbf{(C)}=4+3\sqrt 2}</math>. | ||
| + | |||
| + | == See Also == | ||
| + | |||
| + | {{AMC10 box|year=2002|ab=B|num-b=16|num-a=18}} | ||
| + | |||
| + | [[Category:Introductory Geometry Problems]] | ||
| + | [[Category:Area Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 12:27, 21 May 2021
Problem
A regular octagon 
has sides of length two. Find the area of 
.
Solution
![[asy] unitsize(1cm); defaultpen(0.8); pair[] A = new pair[8]; A[0]=(0,0); for (int i=1; i<8; ++i) A[i] = A[i-1] + 2*dir(45*(i-1)); draw( A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle ); label("$A$",A[0],SW); label("$B$",A[1],SE); label("$C$",A[2],SE); label("$D$",A[3],NE); label("$E$",A[4],NE); label("$F$",A[5],NW); label("$G$",A[6],NW); label("$H$",A[7],SW); filldraw( A[0]--A[3]--A[6]--cycle, lightgray, black ); pair P = intersectionpoint( A[3]--A[6], A[0]--A[5] ); draw( A[0]--P ); draw( P -- A[5], dashed ); label("$P$",P,NE); draw( A[1]--A[4], dashed ); pair Q = intersectionpoint( A[3]--A[6], A[1]--A[4] ); label("$Q$",Q,NW); [/asy]](http://latex.artofproblemsolving.com/c/a/9/ca90d855ac2876463b79d50db160c0d88e6f0f30.png)
The area of the triangle 
can be computed as 
. We will now find 
and 
.
Clearly, 
is a right isosceles triangle with hypotenuse of length 
, hence 
.
The same holds for triangle 
and its leg 
. The length of 
is equal to 
.
Hence 
, and 
.
Then the area of equals 
.
See Also
| 2002 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
