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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>
  
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==Solution==
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==Solutions==
 
===Solution 1===
 
===Solution 1===
  
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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>
  
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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>
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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.
  
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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>.
  
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===Solution 4===
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Find the area of the bottom triangle, which is 4.5. Notice that the area of the triangles is divided by 4 every time. 4.5*5/4≈5.7, and 5.7+(1/4)≈5.9. We can clearly see that the sum is approaching answer choice <math>\boxed{A}</math>.
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===Solution 5===
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Obviously, the area of <math>DBC</math> is <math>\frac{1}{3}</math> of <math>AJDC</math>, the area of <math>DEA</math> is <math>\frac{1}{3}</math> of <math>JIED</math>, if the progress is going to infinity, the shaded triangles will be <math>\frac{1}{3}</math> of the triangle <math>ACG</math>. However, 100 times is much enough. The answer is <math>\frac{1}{3}\times 6\times 6=\boxed{(A)6}</math>.
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 +
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
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 +
https://www.youtube.com/watch?v=sZabsoMIf2I
  
 
==See Also==
 
==See Also==

Latest revision as of 04:21, 12 March 2024

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]

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

Solutions

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 $\boxed{A}$.

Solution 4

Find the area of the bottom triangle, which is 4.5. Notice that the area of the triangles is divided by 4 every time. 4.5*5/4≈5.7, and 5.7+(1/4)≈5.9. We can clearly see that the sum is approaching answer choice $\boxed{A}$.

Solution 5

Obviously, the area of $DBC$ is $\frac{1}{3}$ of $AJDC$, the area of $DEA$ is $\frac{1}{3}$ of $JIED$, if the progress is going to infinity, the shaded triangles will be $\frac{1}{3}$ of the triangle $ACG$. However, 100 times is much enough. The answer is $\frac{1}{3}\times 6\times 6=\boxed{(A)6}$.

Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=sZabsoMIf2I

See Also

1999 AMC 8 (Problems • Answer Key • Resources)
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