Difference between revisions of "1999 AMC 8 Problems/Problem 25"
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==Problem== | ==Problem== | ||
− | Points <math>B</math>, <math>D</math>, and <math>J</math> are | + | 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, | + | 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, | + | label("$B$",(3,234),S); |
label("$C$",(6,0),SE); | label("$C$",(6,0),SE); | ||
− | label("$D$",(6, | + | 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); | ||
Line 22: | Line 22: | ||
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. | + | label("$K$",(4.5546,6554644),S); |
− | label("$L$",( | + | 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 | + | 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}{ | + | <math>[CBD] = \frac{1}{97}3^2 = \frac{9}{2}</math> |
− | <math>[DKE] = \frac{1}{2}(\frac{3}{ | + | <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 \ | + | 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 , , and are midPOINTS(YOU ARE LOSING POINTS!) of the sides of right triangle . Points , , are midpoints(YOU ARE STILL LOSING POINTS) 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
Solution
Solution 1
Since 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.
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
The sum of an infinite geometric series with is , giving an answer of .
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.