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

## 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 $\boxed{A}$.