Difference between revisions of "1999 AMC 8 Problems/Problem 25"
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− | == | + | ==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 | 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 | ||
− | < | + | <asy> |
draw((0,0)--(6,0)--(6,6)--cycle); | draw((0,0)--(6,0)--(6,6)--cycle); | ||
draw((3,0)--(3,3)--(6,3)); | draw((3,0)--(3,3)--(6,3)); | ||
Line 12: | Line 12: | ||
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(" | + | label("$A$",(0,0),SW); |
− | label(" | + | label("$B$",(3,0),S); |
− | label(" | + | label("$C$",(6,0),SE); |
− | label(" | + | label("$D$",(6,3),E); |
− | label(" | + | label("$E$",(6,4.5),E); |
− | label(" | + | label("$F$",(6,5.25),E); |
− | label(" | + | label("$G$",(6,6),NE); |
− | label(" | + | label("$H$",(5.25,5.25),NW); |
− | label(" | + | label("$I$",(4.5,4.5),NW); |
− | label(" | + | label("$J$",(3,3),NW); |
− | label(" | + | label("$K$",(4.5,3),S); |
− | label(" | + | label("$L$",(5.25,4.5),S); |
</asy> | </asy> | ||
<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> | ||
− | == | + | ==Solutions== |
===Solution 1=== | ===Solution 1=== | ||
Line 67: | Line 67: | ||
<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> | ||
− | 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 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. |
− | The sum of an infinite geometric series with <math>|r|<1</math> is <math>S_{\infty} = \frac{a_1}{1 - r} | + | 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>. |
+ | ===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>. | ||
+ | |||
+ | ===Solution 5=== | ||
+ | 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>. | ||
+ | |||
+ | == Video Solution by CosineMethod [🔥Fast and Easy🔥]== | ||
+ | |||
+ | https://www.youtube.com/watch?v=sZabsoMIf2I | ||
==See Also== | ==See Also== |
Latest revision as of 05:21, 12 March 2024
Contents
Problem
Points ,
, and
are midpoints of the sides of right triangle
. Points
,
,
are midpoints of the sides of triangle
, etc. If the dividing and shading process is done 100 times (the first three are shown) and
, then the total area of the shaded triangles is nearest
Solutions
Solution 1
Since 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.
The sum of the shaded regions is
is an underestimate, as some portion (but not all) of
will be shaded in future iterations.
If you shade all of , this will add an additional
to the area, giving
, which is an overestimate.
Thus, is the only answer that is both over the underestimate and under the overestimate.
Solution 2
In iteration , congruent triangles
and
are created, with one of them being shaded.
In iteration , three more congruent triangles are created, with one of them being shaded.
As the process continues indefnitely, in each row, of each triplet of new congruent triangles will be shaded. The "fourth triangle" at the top (
in the diagram) will gradually shrink,
leaving about of the area shaded. This means
square units will be shaded when the process goes on indefinitely, giving
.
Solution 3
Using Solution 1 as a template, note that the sum of the areas forms a geometric series:
This is the sum of a geometric series with first term and common ratio
This is the easiest way to do this problem.
The sum of an infinite geometric series with is shown by the formula.
Insert the values to get
, giving an answer of
.
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 .
Solution 5
Obviously, the area of is
of
, the area of
is
of
, if the progress is going to infinity, the shaded triangles will be
of the triangle
. However, 100 times is much enough. The answer is
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.