Difference between revisions of "1999 AMC 8 Problems/Problem 25"
(→Solution 6 (Sigma)) |
(Removed spam) |
||
| Line 1: | Line 1: | ||
==Problem== | ==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 | ||
| Line 29: | Line 28: | ||
==Solution 1== | ==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. | ||
| Line 51: | Line 49: | ||
==Solution 2== | ==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. | ||
| Line 61: | Line 58: | ||
==Solution 3== | ==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]]: | ||
| Line 76: | Line 72: | ||
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== | |
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>. | ||
~RandomMathGuy500 | ~RandomMathGuy500 | ||
| − | ==Solution 7 | + | ==Solution 7== |
| − | + | 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 9 is \boxed{(A)6}$. | |
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| + | ==Video Solution by CosineMethod [Fast and Easy]== | ||
https://www.youtube.com/watch?v=sZabsoMIf2I | https://www.youtube.com/watch?v=sZabsoMIf2I | ||
==See Also== | ==See Also== | ||
| − | |||
{{AMC8 box|year=1999|num-b=24|after=Last Question}} | {{AMC8 box|year=1999|num-b=24|after=Last Question}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 15:32, 5 January 2025
Contents
Problem
Points 
, 
, and 
are midpoints of the sides of right triangle 
. Points 
, 
, 
are midpoints of the sides of triangle 
, etc. If the dividing and shading process is done 100 times (the first three are shown) and 
, then the total area of the shaded triangles is nearest
![[asy] draw((0,0)--(6,0)--(6,6)--cycle); draw((3,0)--(3,3)--(6,3)); draw((4.5,3)--(4.5,4.5)--(6,4.5)); draw((5.25,4.5)--(5.25,5.25)--(6,5.25)); fill((3,0)--(6,0)--(6,3)--cycle,black); fill((4.5,3)--(6,3)--(6,4.5)--cycle,black); fill((5.25,4.5)--(6,4.5)--(6,5.25)--cycle,black); label("$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$",(5.25,4.5),S); [/asy]](http://latex.artofproblemsolving.com/c/f/d/cfd00c85535b42635e07b9230f8dc34f5d21d985.png)
Solution 1
Since 
is fairly small relative to the rest of the diagram, we can make an underestimate by using the current diagram. All triangles are right-isosceles triangles.
![$[CBD] = \frac{1}{2}3^2 = \frac{9}{2}$](http://latex.artofproblemsolving.com/a/6/d/a6d2257c3fcbe4d4ee86d405baa2688a636b3340.png)
![$[DKE] = \frac{1}{2}(\frac{3}{2})^2 = \frac{9}{8}$](http://latex.artofproblemsolving.com/8/e/b/8eb903ca74d854565553d401612d5be1dffa2dfb.png)
![$[ELF] = \frac{1}{2}(\frac{3}{4})^2 = \frac{9}{32}$](http://latex.artofproblemsolving.com/7/2/4/724d5f8c2f455d72a5550e85024df72b2e1bddf7.png)
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
The area of 
is clearly 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 
.
Solution 6
The formula for a geometric series like this is 
. Using this, we get 
.
~RandomMathGuy500
Solution 7
Notice how you can fit 
into the top right corner of square 
? Repeat this process by putting smaller triangles into the corner of the square and eventually, you should see that this converges to just 
of the square. 
is 
of the square and rest of the triangles make which totals to 
of the square. Square has side length 
so it's area is 
, and of 9 is \boxed{(A)6}$.
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. 
