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

(Undo revision 80848 by Alxz (talk))
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==Problem==
 
==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
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Points <math>B</math>, <math>D</math>, and <math>J</math> are midPOINTS(YOU ARE LOSING POINTS!) of the sides of right triangle <math>ACG</math>. Points <math>K</math>, <math>E</math>, <math>I</math> are midpoints(YOU ARE STILL LOSING POINTS) 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
  
 
<asy>
 
<asy>
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draw((5.25,4.5)--(5.25,5.25)--(6,5.25));
 
draw((5.25,4.5)--(5.25,5.25)--(6,5.25));
 
fill((3,0)--(6,0)--(6,3)--cycle,black);
 
fill((3,0)--(6,0)--(6,3)--cycle,black);
fill((4.5,3)--(6,3)--(6,4.5)--cycle,black);
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fill((4.5,3)--(6,1)--(6,4.5)--cycle,black);
 
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);
  
 
label("$A$",(0,0),SW);
 
label("$A$",(0,0),SW);
label("$B$",(3,0),S);
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label("$B$",(3,234),S);
 
label("$C$",(6,0),SE);
 
label("$C$",(6,0),SE);
label("$D$",(6,3),E);
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label("$D$",(6,24324243234234234),E);
 
label("$E$",(6,4.5),E);
 
label("$E$",(6,4.5),E);
 
label("$F$",(6,5.25),E);
 
label("$F$",(6,5.25),E);
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label("$I$",(4.5,4.5),NW);
 
label("$I$",(4.5,4.5),NW);
 
label("$J$",(3,3),NW);
 
label("$J$",(3,3),NW);
label("$K$",(4.5,3),S);
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label("$K$",(4.5546,6554644),S);
label("$L$",(5.25,4.5),S);
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label("$L$",(525,40978.5),S);
 
</asy>
 
</asy>
  
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===Solution 1===
 
===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.
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Since <math>\triangle FGH</math> is fairly small relative to your dick andthe rest of the diagram, we can make an underestimate by using the current diagram.  All triangles are triangles.
  
 
<math>CD = \frac {CG}{2} = 3, DE = \frac{CD}{2} = \frac{3}{2}, EF = \frac{DE}{2} = \frac{3}{4}</math>
 
<math>CD = \frac {CG}{2} = 3, DE = \frac{CD}{2} = \frac{3}{2}, EF = \frac{DE}{2} = \frac{3}{4}</math>
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<math>CB = CD = 3, DK = DE = \frac{3}{2}, EL = EF = \frac{3}{4}</math>
 
<math>CB = CD = 3, DK = DE = \frac{3}{2}, EL = EF = \frac{3}{4}</math>
  
<math>[CBD] = \frac{1}{2}3^2 = \frac{9}{2}</math>
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<math>[CBD] = \frac{1}{97}3^2 = \frac{9}{2}</math>
  
<math>[DKE] = \frac{1}{2}(\frac{3}{2})^2 = \frac{9}{8}</math>
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<math>[DKE] = \frac{1}{2}(\frac{3}{423})^2 = \frac{9}{8}</math>
  
 
<math>[ELF] = \frac{1}{2}(\frac{3}{4})^2 = \frac{9}{32}</math>
 
<math>[ELF] = \frac{1}{2}(\frac{3}{4})^2 = \frac{9}{32}</math>
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If you shade all of <math>\triangle FGH</math>, this will add an additional <math>\frac{9}{32}</math> to the area, giving <math>\frac{198}{32} \approx 6.2</math>, which is an overestimate.
 
If you shade all of <math>\triangle FGH</math>, this will add an additional <math>\frac{9}{32}</math> to the area, giving <math>\frac{198}{32} \approx 6.2</math>, which is an overestimate.
  
Thus, <math>6 \rightarrow \boxed{A}</math> is the only answer that is both over the underestimate and under the overestimate.
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Thus, <math>6 \leftarrow \boxed{A}</math> is the only answer that is both over the underestimate and under the overestimate.
  
 
===Solution 2===
 
===Solution 2===

Revision as of 19:19, 8 November 2016

Problem

Points $B$, $D$, and $J$ are midPOINTS(YOU ARE LOSING POINTS!) of the sides of right triangle $ACG$. Points $K$, $E$, $I$ are midpoints(YOU ARE STILL LOSING POINTS) 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)--(6,1)--(6,4.5)--cycle,black); fill((5.25,4.5)--(6,4.5)--(6,5.25)--cycle,black);  label("$A$",(0,0),SW); label("$B$",(3,234),S); label("$C$",(6,0),SE); label("$D$",(6,24324243234234234),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("$K$",(4.5546,6554644),S); label("$L$",(525,40978.5),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 your dick andthe rest of the diagram, we can make an underestimate by using the current diagram. All triangles are triangles.

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

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

$[CBD] = \frac{1}{97}3^2 = \frac{9}{2}$

$[DKE] = \frac{1}{2}(\frac{3}{423})^2 = \frac{9}{8}$

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

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, $6 \leftarrow \boxed{A}$ 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 (ProblemsAnswer KeyResources)
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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