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Difference between revisions of "1999 AMC 8 Problems/Problem 25"

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(Undo revision 80848 by Alxz (talk))
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==GUO ZAN DAAAAAAAAAAAAAAAAAA==
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==Problem==
  
 
Points <math>B</math>, <math>D</math>, and <math>J</math> are midpoints of the sides of right triangle <math>ACG</math>. Points <math>K</math>, <math>E</math>, <math>I</math> are midpoints of the sides of triangle <math>JDG</math>, etc. If the dividing and shading process is done 100 times (the first three are shown) and <math>AC=CG=6</math>, then the total area of the shaded triangles is nearest
 
Points <math>B</math>, <math>D</math>, and <math>J</math> are midpoints of the sides of right triangle <math>ACG</math>. Points <math>K</math>, <math>E</math>, <math>I</math> are midpoints of the sides of triangle <math>JDG</math>, etc. If the dividing and shading process is done 100 times (the first three are shown) and <math>AC=CG=6</math>, then the total area of the shaded triangles is nearest
  
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<BABAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA YI GU BABA GUO FEN LAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA>
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<asy>
 
draw((0,0)--(6,0)--(6,6)--cycle);
 
draw((0,0)--(6,0)--(6,6)--cycle);
 
draw((3,0)--(3,3)--(6,3));
 
draw((3,0)--(3,3)--(6,3));
Line 12: Line 12:
 
fill((5.25,4.5)--(6,4.5)--(6,5.25)--cycle,black);
 
fill((5.25,4.5)--(6,4.5)--(6,5.25)--cycle,black);
  
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label("<math>A</math>",(0,0),SW);
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label("$A$",(0,0),SW);
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label("<math>B</math>",(3,0),S);
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label("$B$",(3,0),S);
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label("<math>C</math>",(6,0),SE);
+
label("$C$",(6,0),SE);
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label("<math>D</math>",(6,3),E);
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label("$D$",(6,3),E);
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label("<math>E</math>",(6,4.5),E);
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label("$E$",(6,4.5),E);
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label("<math>F</math>",(6,5.25),E);
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label("$F$",(6,5.25),E);
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label("<math>G</math>",(6,6),NE);
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label("$G$",(6,6),NE);
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label("<math>H</math>",(5.25,5.25),NW);
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label("$H$",(5.25,5.25),NW);
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label("<math>I</math>",(4.5,4.5),NW);
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label("$I$",(4.5,4.5),NW);
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label("<math>J</math>",(3,3),NW);
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label("$J$",(3,3),NW);
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label("<math>K</math>",(4.5,3),S);
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label("$K$",(4.5,3),S);
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label("<math>L</math>",(5.25,4.5),S);
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label("$L$",(5.25,4.5),S);
 
</asy>
 
</asy>
  

Revision as of 15:41, 31 October 2016

Problem

Points $B$, $D$, and $J$ are midpoints of the sides of right triangle $ACG$. Points $K$, $E$, $I$ are midpoints of the sides of triangle $JDG$, etc. If the dividing and shading process is done 100 times (the first three are shown) and $AC=CG=6$, then the total area of the shaded triangles is nearest

[asy] draw((0,0)--(6,0)--(6,6)--cycle); draw((3,0)--(3,3)--(6,3)); draw((4.5,3)--(4.5,4.5)--(6,4.5)); draw((5.25,4.5)--(5.25,5.25)--(6,5.25)); fill((3,0)--(6,0)--(6,3)--cycle,black); fill((4.5,3)--(6,3)--(6,4.5)--cycle,black); fill((5.25,4.5)--(6,4.5)--(6,5.25)--cycle,black); label("$A$",(0,0),SW); label("$B$",(3,0),S); label("$C$",(6,0),SE); label("$D$",(6,3),E); label("$E$",(6,4.5),E); label("$F$",(6,5.25),E); label("$G$",(6,6),NE); label("$H$",(5.25,5.25),NW); label("$I$",(4.5,4.5),NW); label("$J$",(3,3),NW); label("$K$",(4.5,3),S); label("$L$",(5.25,4.5),S); [/asy]

$\text{(A)}\ 6 \qquad \text{(B)}\ 7 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 10$

Solution

Solution 1

Since $\triangle FGH$ is fairly small relative to the rest of the diagram, we can make an underestimate by using the current diagram. All triangles are right-isosceles triangles.

$CD = \frac {CG}{2} = 3, DE = \frac{CD}{2} = \frac{3}{2}, EF = \frac{DE}{2} = \frac{3}{4}$

$CB = CD = 3, DK = DE = \frac{3}{2}, EL = EF = \frac{3}{4}$

$[CBD] = \frac{1}{2}3^2 = \frac{9}{2}$

$[DKE] = \frac{1}{2}(\frac{3}{2})^2 = \frac{9}{8}$

$[ELF] = \frac{1}{2}(\frac{3}{4})^2 = \frac{9}{32}$

The sum of the shaded regions is $\frac{9}{2} + \frac{9}{8} + \frac{9}{32} = \frac{189}{32} \approx 5.9$

$5.9$ is an underestimate, as some portion (but not all) of $\triangle FGH$ will be shaded in future iterations.

If you shade all of $\triangle FGH$, this will add an additional $\frac{9}{32}$ to the area, giving $\frac{198}{32} \approx 6.2$, which is an overestimate.

Thus, $6 \rightarrow \boxed{A}$ is the only answer that is both over the underestimate and under the overestimate.

Solution 2

In iteration $1$, congruent triangles $\triangle ABJ, \triangle BDJ,$ and $\triangle BDC$ are created, with one of them being shaded.

In iteration $2$, three more congruent triangles are created, with one of them being shaded.

As the process continues indefnitely, in each row, $\frac{1}{3}$ of each triplet of new congruent triangles will be shaded. The "fourth triangle" at the top ($\triangle FGH$ in the diagram) will gradually shrink,

leaving about $\frac{1}{3}$ of the area shaded. This means $\frac{1}{3}\left(\frac{1}{2}6\cdot 6\right) = 6$ square units will be shaded when the process goes on indefinitely, giving $\boxed{A}$.

Solution 3

Using Solution 1 as a template, note that the sum of the areas forms a geometric series:

$\frac{9}{2} + \frac{9}{8} + \frac{9}{32} + \frac{9}{128} + ...$

This is the sum of a geometric series with first term $a_1 = \frac{9}{2}$ and common ratio $r = \frac{1}{4}$

The sum of an infinite geometric series with $|r|<1$ is $S_{\infty} = \frac{a_1}{1 - r} = \frac{\frac{9}{2}}{1 - \frac{1}{4}} = \frac{9}{2}\cdot\frac{4}{3} = 6$, giving an answer of $\boxed{A}$.


See Also

1999 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Question
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 AJHSME/AMC 8 Problems and Solutions

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