Difference between revisions of "2015 AMC 8 Problems/Problem 25"
m (→See Also) |
Pi is 3.14 (talk | contribs) (→Video Solution) |
||
(60 intermediate revisions by 35 users not shown) | |||
Line 1: | Line 1: | ||
+ | == Problem == | ||
+ | |||
One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space? | One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space? | ||
− | |||
− | |||
<asy> | <asy> | ||
+ | size(75); | ||
draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); | draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); | ||
filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); | filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); | ||
Line 11: | Line 12: | ||
</asy> | </asy> | ||
+ | <math>\textbf{(A) } 9\qquad \textbf{(B) } 12\frac{1}{2}\qquad \textbf{(C) } 15\qquad \textbf{(D) } 15\frac{1}{2}\qquad \textbf{(E) } 17</math> | ||
+ | |||
+ | ==Solutions== | ||
===Solution 1=== | ===Solution 1=== | ||
− | We can draw a diagram as shown | + | We can draw a diagram as shown: |
<asy> | <asy> | ||
+ | size(75); | ||
draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); | draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); | ||
filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); | filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); | ||
Line 24: | Line 29: | ||
draw(P--Pp--Ppp--Pppp--cycle); | draw(P--Pp--Ppp--Pppp--cycle); | ||
</asy> | </asy> | ||
− | Let us focus on the big triangles | + | Let us focus on the <math>4</math> big triangles that, together with the inscribed square, fill in the large square. If we zoom in on one of the four triangles, we can see that it is composed of a small unit square in the corner of the large square and two triangles, one smaller than the other. We are going to focus specifically on the smaller of the two triangles. This triangle is similar to the big triangle itself by <math>\mathrm{AA}</math> similarity(because the two sides of a square are parallel. To prove this fact, draw a diagonal of the square and find congruent triangles). Let the shorter leg of the big triangle be <math>x</math>; then <math>\dfrac{x}{x-1}=\dfrac{5-x}{1}</math>. |
<cmath>x=-x^2+6x-5</cmath> | <cmath>x=-x^2+6x-5</cmath> | ||
<cmath>x^2-5x+5=0</cmath> | <cmath>x^2-5x+5=0</cmath> | ||
<cmath>x=\dfrac{5\pm \sqrt{(-5)^2-(4)(1)(5)}}{2}</cmath> | <cmath>x=\dfrac{5\pm \sqrt{(-5)^2-(4)(1)(5)}}{2}</cmath> | ||
<cmath>x=\dfrac{5\pm \sqrt{5}}{2}</cmath> | <cmath>x=\dfrac{5\pm \sqrt{5}}{2}</cmath> | ||
− | Thus <math>x=\dfrac{5-\sqrt{5}}{2}</math> | + | Thus <math>x=\dfrac{5-\sqrt{5}}{2}</math>, because by symmetry, <math>x < \dfrac52</math>. Note that the other solution we got, namely, <math>x=\dfrac{5+\sqrt 5} 2</math>, is the length of the segment <math>5-x</math>. <math>x</math> and <math>5-x</math> together sum to <math>5</math>, the side of the length of the large square, and similarly, the sum of the solutions is <math>5</math>. This solution is a result of the symmetry of the problem; if we had set the longer leg of the big triangle to be <math>x</math>, then we would solve the same quadratic to find the same roots, the only difference being that we take the other root. |
+ | |||
This means the area of each triangle is <math>\dfrac{5-\sqrt{5}}{2}*(5-\dfrac{5-\sqrt{5}}{2})*\dfrac{1}{2}=\dfrac{5}{2}</math> | This means the area of each triangle is <math>\dfrac{5-\sqrt{5}}{2}*(5-\dfrac{5-\sqrt{5}}{2})*\dfrac{1}{2}=\dfrac{5}{2}</math> | ||
− | + | Thus the area of the square is <math>25-(4*\dfrac{5}{2})=15\implies \boxed{\textbf{(C)}}</math> | |
+ | |||
+ | Don't use this in a contest scenario, only use when practicing math skills :) | ||
− | ===Solution 2=== | + | ===Solution 2 (Contest Solution)=== |
We draw a square as shown: | We draw a square as shown: | ||
+ | |||
<asy> | <asy> | ||
− | + | size(75); | |
− | + | draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); | |
− | + | filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); | |
− | + | filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); | |
− | + | filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); | |
− | draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); | + | filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); |
− | filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); | + | path arc = arc((2.5,4),1.5,0,90); |
− | filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); | + | pair P = intersectionpoint(arc,(0,5)--(5,5)); |
− | filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); | + | pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp; |
− | filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); | + | draw(P--Pp--Ppp--Pppp--cycle); |
− | draw( | + | filldraw((1,4)--P--(4,4)--cycle,red); |
− | + | filldraw((4,4)--Pppp--(4,1)--cycle,red); | |
+ | filldraw((1,1)--Ppp--(4,1)--cycle,red); | ||
+ | filldraw((1,1)--Pp--(1,4)--cycle,red); | ||
</asy> | </asy> | ||
− | We wish to find the area of the square. | + | |
+ | We wish to find the area of the square. The area of the larger square is composed of the smaller square and the four red triangles. Because the inner square has an area of <math>(5-2)3</math> so it has length <math>\sqrt{9}=3</math> and because the height of the red triangle is 1(because the gray squares have length one), the area of one triangle is namely <math>\frac{3 \cdot 1}{2}</math>. Thus, the combined area of the four triangles is <math>4 \cdot \frac 32=6</math>. Furthermore, the area of the smaller square is <math>9</math>. We add these to see that the area of the large square is <math>9+6=15\implies \boxed{\textbf{(C)}}</math>. | ||
===Solution 3=== | ===Solution 3=== | ||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | <math> | + | Let us find the area of the triangles and the unit squares: on each side, there are two triangles. They both have one leg of length <math>1</math>, and let's label the other legs <math>x</math> for one of the triangles and <math>y</math> for the other. Note that <math>x + y = 3</math>. |
− | <math> | + | The area of each of the triangles is <math>\frac{x}{2}</math> and <math>\frac{y}{2}</math>, and there are <math>4</math> of each. So now we need to find <math>4\left(\frac{x}{2}\right) + 4\left(\frac{y}{2}\right)</math>. Expanding out the fractions, we get <math>4\left(\frac{x}{2}\right) + 4\left(\frac{y}{2}\right) = 2x + 2y = 2(x + y) = 2(3) = 6</math>. |
− | <math> | + | So the area of the square we need is <math>25- (4+6) = 15\implies \boxed{\textbf{(C)}}</math>. |
− | + | ||
− | + | ==Video Solution by OmegaLearn== | |
+ | https://youtu.be/51K3uCzntWs?t=3358 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/WNiJWmKCfj0 - Happytwin | ||
+ | |||
+ | https://www.youtube.com/watch?v=_YAnHtQJ5qA | ||
+ | |||
+ | https://youtu.be/rTljQV79PCY | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== | ||
− | {{AMC8 box|year=2015|num-b=24|after=}} | + | {{AMC8 box|year=2015|num-b=24|after=Last Problem}} |
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 19:55, 2 January 2023
Contents
Problem
One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space?
Solutions
Solution 1
We can draw a diagram as shown: Let us focus on the big triangles that, together with the inscribed square, fill in the large square. If we zoom in on one of the four triangles, we can see that it is composed of a small unit square in the corner of the large square and two triangles, one smaller than the other. We are going to focus specifically on the smaller of the two triangles. This triangle is similar to the big triangle itself by similarity(because the two sides of a square are parallel. To prove this fact, draw a diagonal of the square and find congruent triangles). Let the shorter leg of the big triangle be ; then . Thus , because by symmetry, . Note that the other solution we got, namely, , is the length of the segment . and together sum to , the side of the length of the large square, and similarly, the sum of the solutions is . This solution is a result of the symmetry of the problem; if we had set the longer leg of the big triangle to be , then we would solve the same quadratic to find the same roots, the only difference being that we take the other root.
This means the area of each triangle is Thus the area of the square is
Don't use this in a contest scenario, only use when practicing math skills :)
Solution 2 (Contest Solution)
We draw a square as shown:
We wish to find the area of the square. The area of the larger square is composed of the smaller square and the four red triangles. Because the inner square has an area of so it has length and because the height of the red triangle is 1(because the gray squares have length one), the area of one triangle is namely . Thus, the combined area of the four triangles is . Furthermore, the area of the smaller square is . We add these to see that the area of the large square is .
Solution 3
Let us find the area of the triangles and the unit squares: on each side, there are two triangles. They both have one leg of length , and let's label the other legs for one of the triangles and for the other. Note that . The area of each of the triangles is and , and there are of each. So now we need to find . Expanding out the fractions, we get . So the area of the square we need is .
Video Solution by OmegaLearn
https://youtu.be/51K3uCzntWs?t=3358
~ pi_is_3.14
Video Solution
https://youtu.be/WNiJWmKCfj0 - Happytwin
https://www.youtube.com/watch?v=_YAnHtQJ5qA
~savannahsolver
See Also
2015 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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.