Difference between revisions of "2007 AMC 10A Problems/Problem 18"
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+ | ==Problem== | ||
+ | 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> | ||
+ | |||
+ | ==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 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | === 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=(12/5,24/5).</math> Using [[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> | ||
+ | |||
+ | ==See also== | ||
+ | {{AMC10 box|year=2007|ab=A|num-b=17|num-a=19}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 14:20, 20 October 2018
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 Shoelace Theorem (or you could connect and solve for the resulting triangle + trapezoid areas), we find
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.