Difference between revisions of "2007 AMC 10A Problems/Problem 18"
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Consider the <math>12</math>-sided polygon <math>ABCDEFGHIJKL</math>, as shown. Each of its sides has length <math>4</math>, and each two consecutive sides form a right angle. Suppose that <math>\overline{AG}</math> and <math>\overline{CH}</math> meet at <math>M</math>. What is the area of quadrilateral <math>ABCM</math>? | Consider the <math>12</math>-sided polygon <math>ABCDEFGHIJKL</math>, as shown. Each of its sides has length <math>4</math>, and each two consecutive sides form a right angle. Suppose that <math>\overline{AG}</math> and <math>\overline{CH}</math> meet at <math>M</math>. What is the area of quadrilateral <math>ABCM</math>? | ||
− | [ | + | <asy> |
+ | unitsize(13mm); | ||
+ | defaultpen(linewidth(.8pt)+fontsize(10pt)); | ||
+ | dotfactor=4; | ||
+ | |||
+ | pair A=(1,3), B=(2,3), C=(2,2), D=(3,2), Ep=(3,1), F=(2,1), G=(2,0), H=(1,0), I=(1,1), J=(0,1), K=(0,2), L=(1,2); | ||
+ | pair M=intersectionpoints(A--G,H--C)[0]; | ||
+ | |||
+ | draw(A--B--C--D--Ep--F--G--H--I--J--K--L--cycle); | ||
+ | draw(A--G); | ||
+ | draw(H--C); | ||
+ | dot(M); | ||
+ | |||
+ | label("$A$",A,NW); | ||
+ | label("$B$",B,NE); | ||
+ | label("$C$",C,NE); | ||
+ | label("$D$",D,NE); | ||
+ | label("$E$",Ep,SE); | ||
+ | label("$F$",F,SE); | ||
+ | label("$G$",G,SE); | ||
+ | label("$H$",H,SW); | ||
+ | label("$I$",I,SW); | ||
+ | label("$J$",J,SW); | ||
+ | label("$K$",K,NW); | ||
+ | label("$L$",L,NW); | ||
+ | label("$M$",M,W); | ||
+ | </asy> | ||
<math>\text{(A)}\ \frac {44}{3}\qquad \text{(B)}\ 16 \qquad \text{(C)}\ \frac {88}{5}\qquad \text{(D)}\ 20 \qquad \text{(E)}\ \frac {62}{3}</math> | <math>\text{(A)}\ \frac {44}{3}\qquad \text{(B)}\ 16 \qquad \text{(C)}\ \frac {88}{5}\qquad \text{(D)}\ 20 \qquad \text{(E)}\ \frac {62}{3}</math> | ||
==Solution== | ==Solution== | ||
+ | |||
+ | === Solution 1 === | ||
We can obtain the solution by calculating the area of rectangle <math>ABGH</math> minus the combined area of triangles <math>\triangle AHG</math> and <math>\triangle CGM</math>. | We can obtain the solution by calculating the area of rectangle <math>ABGH</math> minus the combined area of triangles <math>\triangle AHG</math> and <math>\triangle CGM</math>. | ||
− | We know that triangles <math>\triangle AMH</math> and <math>\triangle | + | We know that triangles <math>\triangle AMH</math> and <math>\triangle GMC</math> are similar because <math>\overline{AH} \parallel \overline{CG}</math>. Also, since <math>\frac{AH}{CG} = \frac{3}{2}</math>, the ratio of the distance from <math>M</math> to <math>\overline{AH}</math> to the distance from <math>M</math> to <math>\overline{CG}</math> is also <math>\frac{3}{2}</math>. Solving with the fact that the distance from <math>\overline{AH}</math> to <math>\overline{CG}</math> is 4, we see that the distance from <math>M</math> to <math>\overline{CG}</math> is <math>\frac{8}{5}</math>. |
The area of <math>\triangle AHG</math> is simply <math>\frac{1}{2} \cdot 4 \cdot 12 = 24</math>, the area of <math>\triangle CGM</math> is <math>\frac{1}{2} \cdot \frac{8}{5} \cdot 8 = \frac{32}{5}</math>, and the area of rectangle <math>ABGH</math> is <math>4 \cdot 12 = 48</math>. | The area of <math>\triangle AHG</math> is simply <math>\frac{1}{2} \cdot 4 \cdot 12 = 24</math>, the area of <math>\triangle CGM</math> is <math>\frac{1}{2} \cdot \frac{8}{5} \cdot 8 = \frac{32}{5}</math>, and the area of rectangle <math>ABGH</math> is <math>4 \cdot 12 = 48</math>. | ||
− | Taking the area of rectangle <math>ABGH</math> and subtracting the combined area of <math>\triangle AHG</math> and <math>\triangle CGM</math> yields <math>48 - (24 + \frac{32}{5}) = \boxed{\frac{88}{5}}\ \text{(C)}</math>. | + | Taking the area of rectangle <math>ABGH</math> and subtracting the combined area of <math>\triangle AHG</math> and <math>\triangle CGM</math> yields <math>48 - \left(24 + \frac{32}{5}\right) = \boxed{\frac{88}{5}}\ \text{(C)}</math>. |
+ | |||
+ | === Solution 2 === | ||
+ | |||
+ | <asy> | ||
+ | unitsize(2cm); | ||
+ | defaultpen(linewidth(.8pt)+fontsize(10pt)); | ||
+ | dotfactor=4; | ||
+ | |||
+ | pair A=(1,3), B=(2,3), C=(2,2), D=(3,2), Ep=(3,1), F=(2,1), G=(2,0), H=(1,0), I=(1,1), J=(0,1), K=(0,2), L=(1,2); | ||
+ | pair M=intersectionpoints(A--G,H--C)[0]; | ||
+ | pair Z=(2.5,3); | ||
+ | |||
+ | draw(A--B--C--D--Ep--F--G--H--I--J--K--L--cycle); | ||
+ | draw(A--G); | ||
+ | draw(H--C); | ||
+ | draw(B--Z--C); | ||
+ | draw(C--F); | ||
+ | dot(M); | ||
+ | |||
+ | label("$A$",A,NW); | ||
+ | label("$B$",B,N); | ||
+ | label("$C$",C,SE); | ||
+ | label("$D$",D,NE); | ||
+ | label("$E$",Ep,SE); | ||
+ | label("$F$",F,SE); | ||
+ | label("$G$",G,SE); | ||
+ | label("$H$",H,SW); | ||
+ | label("$I$",I,SW); | ||
+ | label("$J$",J,SW); | ||
+ | label("$K$",K,NW); | ||
+ | label("$L$",L,NW); | ||
+ | label("$M$",M,W); | ||
+ | label("$N$",Z,NE); | ||
+ | </asy> | ||
+ | |||
+ | Extend <math>AB</math> and <math>CH</math> and call their intersection <math>N</math>. | ||
+ | |||
+ | The triangles <math>CBN</math> and <math>CGH</math> are clearly similar with ratio <math>1:2</math>, hence <math>BN=2</math> and thus <math>AN=6</math>. The area of the triangle <math>BCN</math> is <math>\frac{2\cdot 4}2 = 4</math>. | ||
+ | |||
+ | The triangles <math>MAN</math> and <math>MGH</math> are similar as well, and we now know that the ratio of their dimensions is <math>AN:GH = 6:4 = 3:2</math>. | ||
+ | |||
+ | Draw altitudes from <math>M</math> onto <math>AN</math> and <math>GH</math>, let their feet be <math>M_1</math> and <math>M_2</math>. We get that <math>MM_1 : MM_2 = 3:2</math>. Hence <math>MM_1 = \frac 35 \cdot 12 = \frac {36}5 </math>. (An alternate way is by seeing that the set-up AHGCM is similar to the 2 pole problem (http://www.artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_30). Therefore, <math>MM_2</math> must be <math>\frac{1}{\frac{1}{8}+\frac{1}{12}} = \frac{24}{5}</math>, by the harmonic mean. Thus, <math>MM_1</math> must be <math>\frac{36}{5}</math>.) | ||
+ | |||
+ | Then the area of <math>AMN</math> is <math>\frac 12 \cdot AN \cdot MM_1 = \frac{108}5</math>, and the area of <math>ABCM</math> can be obtained by subtracting the area of <math>BCN</math>, which is <math>4</math>. Hence the answer is <math>\frac{108}5 - 4 = \boxed{\frac{88}5}</math>. | ||
+ | |||
+ | == Solution 3 == | ||
+ | |||
+ | |||
+ | <asy> | ||
+ | unitsize(13mm); | ||
+ | defaultpen(linewidth(.8pt)+fontsize(10pt)); | ||
+ | dotfactor=4; | ||
+ | |||
+ | pair A=(1,3), B=(2,3), C=(2,2), D=(3,2), Ep=(3,1), F=(2,1), G=(2,0), H=(1,0), I=(1,1), J=(0,1), K=(0,2), L=(1,2); | ||
+ | pair M=intersectionpoints(A--G,H--C)[0]; | ||
+ | |||
+ | draw(A--B--C--D--Ep--F--G--H--I--J--K--L--cycle); | ||
+ | draw(A--G); | ||
+ | draw(H--C); | ||
+ | dot(M); | ||
+ | |||
+ | label("$A$",A,NW); | ||
+ | label("$B$",B,NE); | ||
+ | label("$C$",C,NE); | ||
+ | label("$D$",D,NE); | ||
+ | label("$E$",Ep,SE); | ||
+ | label("$F$",F,SE); | ||
+ | label("$G$",G,SE); | ||
+ | label("$H$",H,SW); | ||
+ | label("$I$",I,SW); | ||
+ | label("$J$",J,SW); | ||
+ | label("$K$",K,NW); | ||
+ | label("$L$",L,NW); | ||
+ | label("$M$",M,W); | ||
+ | </asy> | ||
+ | |||
+ | We can use coordinates to solve this. Let <math>H=(0,0).</math> Thus, we have <math>A=(0,12), C=(4,8), G=(4,0).</math> Therefore, <math>AG</math> has equation <math>-3x+12=y</math> and <math>HC</math> has equation <math>2x=y.</math> Solving, we have <math>M=(\frac {12}{5},\frac {24}{5}).</math> Using the [[Shoelace Theorem]] (or you could connect <math>LC</math> and solve for the resulting triangle + trapezoid areas), we find <math>[ABCM]=\boxed{\mathrm{(C) \ }\dfrac{88}{5}}.</math> | ||
+ | |||
+ | ~(minor edits by Arcticturn) | ||
==See also== | ==See also== | ||
Line 20: | Line 127: | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 00:17, 20 May 2023
Problem
Consider the -sided polygon
, as shown. Each of its sides has length
, and each two consecutive sides form a right angle. Suppose that
and
meet at
. What is the area of quadrilateral
?
Solution
Solution 1
We can obtain the solution by calculating the area of rectangle minus the combined area of triangles
and
.
We know that triangles and
are similar because
. Also, since
, the ratio of the distance from
to
to the distance from
to
is also
. Solving with the fact that the distance from
to
is 4, we see that the distance from
to
is
.
The area of is simply
, the area of
is
, and the area of rectangle
is
.
Taking the area of rectangle and subtracting the combined area of
and
yields
.
Solution 2
Extend and
and call their intersection
.
The triangles and
are clearly similar with ratio
, hence
and thus
. The area of the triangle
is
.
The triangles and
are similar as well, and we now know that the ratio of their dimensions is
.
Draw altitudes from onto
and
, let their feet be
and
. We get that
. Hence
. (An alternate way is by seeing that the set-up AHGCM is similar to the 2 pole problem (http://www.artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_30). Therefore,
must be
, by the harmonic mean. Thus,
must be
.)
Then the area of is
, and the area of
can be obtained by subtracting the area of
, which is
. Hence the answer is
.
Solution 3
We can use coordinates to solve this. Let Thus, we have
Therefore,
has equation
and
has equation
Solving, we have
Using the Shoelace Theorem (or you could connect
and solve for the resulting triangle + trapezoid areas), we find
~(minor edits by Arcticturn)
See also
2007 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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.