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Difference between revisions of "1999 AMC 8 Problems/Problem 25"
(→‎Solution 6 (Sigma))
 
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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>.
  
−
==Solution 6 (Sigma)==
+
fg
−
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>.
 
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>.
  
Line 84: Line 82:
  
 
==Solution 7 (Even more Sigma)==
 
==Solution 7 (Even more Sigma)==
−
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} 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 {9} is \boxed{(A)6}</math>.
+
FEIN FEIN IN THE RIZZ SKIBIDI SIGMA OHIO GYATT LITTLE RIZZLER.
−
 
+
Gyatt
−
~RandomMathGuy500 again  
+
I was in Ohio before I met you
 +
I rizzed too much and that's an issue
 +
But I'm Grimace Shake
 +
Gyatt
 +
Tell your friends it was nice to rizz them
 +
But I hope I never edge again
 +
I know it breaks your Fanum
 +
Taxin' in Ohio and I'm still not sigma
 +
Four years no Livvy
 +
Now you're looking pretty on Adin Ross' Twitchy
 +
And I-I-I-I-I can't rizz
 +
No, I-I-I-I-I can't mew
 +
So Baby Gronk me closer
 +
In the back skibidi toilet
 +
That I know you can't afford
 +
Kai Cenat tatted on my shoulder
 +
Pull the gyatt right off the corner
 +
From that fanum that you taxed
 +
From your roommate back in Ohio
 +
We ain't ever not the rizzler
 +
(Drop: Jelly House)
 +
We ain't ever not the rizzler
 +
We ain't ever not the rizzler
 +
You
 +
Rizz as good as the day I met you
 +
I forget just why I edged you
 +
I was insane
 +
Slay
 +
And play that blink-180 new song
 +
That we gooned to death in Ohio, okay?
 +
I know you love this gyatt
 +
I moved to Ohio in a brainrot car, and
 +
Four years no Griddy
 +
Now I'm lookin' Livvy and you're not skibidi
 +
And I-I-I-I-I can't rizz
 +
No, I-I-I-I-I can't mew
 +
So Baby Gronk me closer
 +
In the back skibidi toilet
 +
That I know you can't afford
 +
Kai Cenat tatted on my shoulder
 +
Pull the gyatt right off the corner
 +
From that fanum that you taxed
 +
From your roommate back in Ohio
 +
We ain't ever not the rizzler
 +
We ain't ever not the rizzler
 +
We ain't ever not the rizzler
 +
So Baby Gronk me closer
 +
In the back skibidi toilet
 +
That I know you can't afford
 +
Kai Cenat tatted on my shoulder
 +
Pull the gyatt right off the corner
 +
From that fanum that you taxed
 +
From your roommate back in Ohio
 +
We ain't ever not the rizzler
 +
We ain't ever not the rizzler
 +
Oh, we ain't ever not the rizzler
 +
We ain't ever not the rizzler
 +
Oh, we ain't ever not the rizzler
 +
We ain't ever not the rizzler
 +
Oh, we ain't ever not the rizzler
 +
We ain't ever not the rizzler
 +
Oh, we ain't ever not the rizzler
 +
We ain't ever not the rizzler
 +
No, we ain't ever not the rizzler
 +
~RandomMathGuy500 again
  
 
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
 
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==

Latest revision as of 04:14, 5 January 2025

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 $\frac{1}{3}\times 6\times 6\times \frac{1}{2} = \boxed{(A)6}$.

fg 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)

FEIN FEIN IN THE RIZZ SKIBIDI SIGMA OHIO GYATT LITTLE RIZZLER. Gyatt I was in Ohio before I met you I rizzed too much and that's an issue But I'm Grimace Shake Gyatt Tell your friends it was nice to rizz them But I hope I never edge again I know it breaks your Fanum Taxin' in Ohio and I'm still not sigma Four years no Livvy Now you're looking pretty on Adin Ross' Twitchy And I-I-I-I-I can't rizz No, I-I-I-I-I can't mew So Baby Gronk me closer In the back skibidi toilet That I know you can't afford Kai Cenat tatted on my shoulder Pull the gyatt right off the corner From that fanum that you taxed From your roommate back in Ohio We ain't ever not the rizzler (Drop: Jelly House) We ain't ever not the rizzler We ain't ever not the rizzler You Rizz as good as the day I met you I forget just why I edged you I was insane Slay And play that blink-180 new song That we gooned to death in Ohio, okay? I know you love this gyatt I moved to Ohio in a brainrot car, and Four years no Griddy Now I'm lookin' Livvy and you're not skibidi And I-I-I-I-I can't rizz No, I-I-I-I-I can't mew So Baby Gronk me closer In the back skibidi toilet That I know you can't afford Kai Cenat tatted on my shoulder Pull the gyatt right off the corner From that fanum that you taxed From your roommate back in Ohio We ain't ever not the rizzler We ain't ever not the rizzler We ain't ever not the rizzler So Baby Gronk me closer In the back skibidi toilet That I know you can't afford Kai Cenat tatted on my shoulder Pull the gyatt right off the corner From that fanum that you taxed From your roommate back in Ohio We ain't ever not the rizzler We ain't ever not the rizzler Oh, we ain't ever not the rizzler We ain't ever not the rizzler Oh, we ain't ever not the rizzler We ain't ever not the rizzler Oh, we ain't ever not the rizzler We ain't ever not the rizzler Oh, we ain't ever not the rizzler We ain't ever not the rizzler No, we ain't ever not the rizzler ~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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