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

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<math>\text{(A)}\ 6 \qquad \text{(B)}\ 7 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 10</math>
 
<math>\text{(A)}\ 6 \qquad \text{(B)}\ 7 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 10</math>
  
==Solution 1==
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==Solution==
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===Solution 1===
  
 
Since <math>\triangle FGH</math> 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.
 
Since <math>\triangle FGH</math> 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.
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Thus, <math>6 \rightarrow \boxed{A}</math> is the only answer that is both over the underestimate and under the overestimate.
 
Thus, <math>6 \rightarrow \boxed{A}</math> is the only answer that is both over the underestimate and under the overestimate.
  
==Solution 2==
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===Solution 2===
  
 
In iteration <math>1</math>, congruent triangles <math>\triangle ABJ,  \triangle BDJ,</math> and <math>\triangle BDC</math> are created, with one of them being shaded.
 
In iteration <math>1</math>, congruent triangles <math>\triangle ABJ,  \triangle BDJ,</math> and <math>\triangle BDC</math> are created, with one of them being shaded.
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leaving about <math>\frac{1}{3}</math> of the area shaded.  This means <math>\frac{1}{3}\left(\frac{1}{2}6\cdot 6\right) = 6</math> square units will be shaded when the process goes on indefinitely, giving <math>\boxed{A}</math>.
 
leaving about <math>\frac{1}{3}</math> of the area shaded.  This means <math>\frac{1}{3}\left(\frac{1}{2}6\cdot 6\right) = 6</math> square units will be shaded when the process goes on indefinitely, giving <math>\boxed{A}</math>.
  
==Solution 3==
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===Solution 3===
  
 
Using Solution 1 as a template, note that the sum of the areas forms a [[geometric series]]:
 
Using Solution 1 as a template, note that the sum of the areas forms a [[geometric series]]:
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<math>\frac{9}{2} + \frac{9}{8} + \frac{9}{32} + \frac{9}{128} + ...</math>
 
<math>\frac{9}{2} + \frac{9}{8} + \frac{9}{32} + \frac{9}{128} + ...</math>
  
This is the sum of a geometric series with first term <math>a_1 = \frac{9}{2}</math> and common ratio <math>r = \frac{1}{4}</math>
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This is the sum of a geometric series with first term <math>a_1 = \frac{9}{2}</math> and common ratio <math>r = \frac{1}{4}</math> This is the easiest way to do this problem.
 
 
The sum of an infinite geometric series with <math>|r|<1</math> is <math>S_{\infty} = \frac{a_1}{1 - r} = \frac{\frac{9}{2}}{1 - \frac{1}{4}} = \frac{9}{2}\cdot\frac{4}{3} = 6</math>, giving an answer of <math>\boxed{A}</math>.
 
  
 +
The sum of an infinite geometric series with <math>|r|<1</math> is shown by the formula. <math>S_{\infty} = \frac{a_1}{1 - r}</math> Insert the values to get <math>\frac{\frac{9}{2}}{1 - \frac{1}{4}} = \frac{9}{2}\cdot\frac{4}{3} = 6</math>, giving an answer of <math>\boxed{A}</math>.
  
 
==See Also==
 
==See Also==
  
{{AMC8 box|year=1999|before=24|after=Last Question}}
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{{AMC8 box|year=1999|num-b=24|after=Last Question}}
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{{MAA Notice}}

Revision as of 13:43, 3 November 2018

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 $r = \frac{1}{4}$ This is the easiest way to do this problem.

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

See Also

1999 AMC 8 (ProblemsAnswer KeyResources)
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Problem 24
Followed by
Last Question
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