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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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==Solutions==
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==Solution 1==
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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.
  
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===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>.
  
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===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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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>.
 
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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==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 <math>\boxed{A}</math>.
 
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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==Solution 5==
 
The area of <math>DBC</math> is clearly <math>\frac{1}{3}</math> of <math>AJDC</math>, the area of <math>DEK</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\times \frac{1}{2} = \boxed{(A)6}</math>.
 
The area of <math>DBC</math> is clearly <math>\frac{1}{3}</math> of <math>AJDC</math>, the area of <math>DEK</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\times \frac{1}{2} = \boxed{(A)6}</math>.
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 +
==Solution 6 (Sigma)==
 +
Because all of the triangles are isosceles, this gives us the area of one triangle with side length <math>s</math>, <math>A = \frac{1}{2}s^2</math>. Logically, <math>s</math> keeps on getting <math>\frac{1}{2}</math> as small as the previous one (<math>3, \frac{3}{2}, \frac{3}{4}</math>, etc) Meaning for triangle number <math>n, s = 3(\frac{1}{2})^n</math>. Putting this into our original equation, <math>A = \frac{1}{2}(3(\frac{1}{2})^n)^2 = \frac{9}{2}(\frac{1}{2})^{2n} = 9(\frac{1}{2})^{2n+1}</math> This means the total area is <math>9\sum^{\infty}_{n=1} \frac{1}{2}^{2n+1}</math>, using exponent laws, we can simplify this to <math>\frac{9}{2} \sum^{\infty}_{n=1} \frac{1}{4}^n</math>.
 +
 +
The formula for a geometric series like this is <math>\sum^{\infty}_{n=1} a*r^n = \frac{a}{1-r}</math>. Using this, we get <math>\frac{\frac{9}{2}}{\frac{4}{4}-\frac{1}{4}} = \frac{9}{2}\times\frac{4}{3} = 3\times2 = \boxed{(A)6}</math>.
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 +
~RandomMathGuy500
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==Solution 7 (Even more Sigma)==
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Notice how you can fit <math>\triangle KDE</math> into the top right corner of square <math>BCDJ</math>? Repeat this process by putting smaller triangles into the corner of the square and eventually, you should see that this converges to just <math>\frac{1}{4}</math> of the square. <math>\triangle BCD</math> is <math>\frac{1}{2}</math> of the square and rest of the triangles make <math>\frac{1}{4}</math> which totals to <math>\frac{3}{4}</math> of the square. Square <math>BCDJ</math> has side length <math>3</math> so it's area is <math>9</math>, and <math>\frac{3}{4}</math> of <math>9</math> is <math>\boxed{(A)6}</math>.
 +
 +
~RandomMathGuy500 again
  
 
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
 
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==

Revision as of 23:59, 30 December 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]

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

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

The area of $DBC$ is clearly $\frac{1}{3}$ of $AJDC$, the area of $DEK$ 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 $\frac{1}{3}\times 6\times 6\times \frac{1}{2} = \boxed{(A)6}$.

Solution 6 (Sigma)

Because all of the triangles are isosceles, this gives us the area of one triangle with side length $s$, $A = \frac{1}{2}s^2$. Logically, $s$ keeps on getting $\frac{1}{2}$ as small as the previous one ($3, \frac{3}{2}, \frac{3}{4}$, etc) Meaning for triangle number $n, s = 3(\frac{1}{2})^n$. Putting this into our original equation, $A = \frac{1}{2}(3(\frac{1}{2})^n)^2 = \frac{9}{2}(\frac{1}{2})^{2n} = 9(\frac{1}{2})^{2n+1}$ This means the total area is $9\sum^{\infty}_{n=1} \frac{1}{2}^{2n+1}$, using exponent laws, we can simplify this to $\frac{9}{2} \sum^{\infty}_{n=1} \frac{1}{4}^n$.

The formula for a geometric series like this is $\sum^{\infty}_{n=1} a*r^n = \frac{a}{1-r}$. Using this, we get $\frac{\frac{9}{2}}{\frac{4}{4}-\frac{1}{4}} = \frac{9}{2}\times\frac{4}{3} = 3\times2 = \boxed{(A)6}$.

~RandomMathGuy500

Solution 7 (Even more Sigma)

Notice how you can fit $\triangle KDE$ into the top right corner of square $BCDJ$? Repeat this process by putting smaller triangles into the corner of the square and eventually, you should see that this converges to just $\frac{1}{4}$ of the square. $\triangle BCD$ is $\frac{1}{2}$ of the square and rest of the triangles make $\frac{1}{4}$ which totals to $\frac{3}{4}$ of the square. Square $BCDJ$ has side length $3$ so it's area is $9$, and $\frac{3}{4}$ of $9$ is $\boxed{(A)6}$.

~RandomMathGuy500 again

Video Solution by CosineMethod [🔥Fast and Easy🔥]

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

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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