Difference between revisions of "2002 AMC 10B Problems/Problem 17"

(Solution)
(6 intermediate revisions by 5 users not shown)
Line 34: Line 34:
 
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>.
 
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 lenght <math>2</math>, hence <math>PG=\sqrt 2</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>.
 
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>.
 
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{4+3\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 ==
 
== See Also ==
  
 
{{AMC10 box|year=2002|ab=B|num-b=16|num-a=18}}
 
{{AMC10 box|year=2002|ab=B|num-b=16|num-a=18}}
 +
 +
[[Category:Introductory Geometry Problems]]
 +
[[Category:Area Problems]]
 +
{{MAA Notice}}

Revision as of 23:42, 23 November 2020

Problem

A regular octagon $ABCDEFGH$ has sides of length two. Find the area of $\triangle ADG$.

$\textbf{(A) } 4 + 2\sqrt2 \qquad \textbf{(B) } 6 + \sqrt2\qquad \textbf{(C) } 4 + 3\sqrt2 \qquad \textbf{(D) } 3 + 4\sqrt2 \qquad \textbf{(E) } 8 + \sqrt2$

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 $ADG$ can be computed as $\frac{DG \cdot AP}2$. We will now find $DG$ and $AP$.

Clearly, $PFG$ is a right isosceles triangle with hypotenuse of length $2$, hence $PG=\sqrt 2$. The same holds for triangle $QED$ and its leg $QD$. The length of $PQ$ is equal to $FE=2$. Hence $GD = 2 + 2\sqrt 2$, and $AP = PD = 2 + \sqrt 2$.

Then the area of $ADG$ equals $\frac{DG \cdot AP}2 = \frac{(2+2\sqrt 2)(2+\sqrt 2)}2 = \frac{8+6\sqrt 2}2 = \boxed{\textbf{(C)}=4+3\sqrt 2}$.

See Also

2002 AMC 10B (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS