Difference between revisions of "2015 AMC 8 Problems/Problem 25"
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− | The left side of the tilted square can be drawn with the equation <math>y=mx+b</math> because it is a straight line segment. The left side also intersects point <math>(1, 1)</math> | + | The left side of the tilted square can be drawn with the equation <math>y=mx+b</math> because it is a straight line segment. The left side also intersects point <math>(1, 1)</math> so if you substitute <math>1</math> for <math>x</math> and <math>y</math> you get the equation <math>1=m+b</math>. <br> |
+ | <br>Similarly, the right side of the tilted square can be drawn with the equation <math>y=mx+b</math> because it is a straight line segment. The right side intersects point <math>(4, 4)</math> so if you substitute <math>4</math> for <math>x</math> and <math>y</math> you get the equation <math>4=4m+b</math> | ||
==See Also== | ==See Also== |
Revision as of 21:49, 11 November 2019
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?
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
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. The red triangles have base and height
, so the combined area of the four triangles is
. 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
.
Remember that
, so substituting this in we find that the area of all of the triangles is
.
The area of the
unit squares is
, so the area of the square we need is
Solution 4
Pretend that the following diagram is on a coordinate plane:
The left side of the tilted square can be drawn with the equation because it is a straight line segment. The left side also intersects point
so if you substitute
for
and
you get the equation
.
Similarly, the right side of the tilted square can be drawn with the equation because it is a straight line segment. The right side intersects point
so if you substitute
for
and
you get the equation
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.