Difference between revisions of "2018 AMC 10A Problems/Problem 23"
m (→Solution 3: clarify possible solutions) |
MRENTHUSIASM (talk | contribs) m (→Video Solution by Richard Rusczyk) |
||
Line 138: | Line 138: | ||
~ dolphin7 | ~ dolphin7 | ||
− | |||
==See Also== | ==See Also== |
Revision as of 11:14, 14 August 2021
- The following problem is from both the 2018 AMC 12A #17 and 2018 AMC 10A #23, so both problems redirect to this page.
Contents
Problem
Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths 3 and 4 units. In the corner where those sides meet at a right angle, he leaves a small unplanted square so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from
to the hypotenuse is 2 units. What fraction of the field is planted?
Solution 1
Let the square have side length . Connect the upper-right vertex of square
with the two vertices of the triangle's hypotenuse. This divides the triangle in several regions whose areas must add up to the area of the whole triangle, which is
.
Square has area
, and the two thin triangle regions have area
and
. The final triangular region with the hypotenuse as its base and height
has area
. Thus, we have
Solving gives . The area of
is
and the desired ratio is
.
Alternatively, once you get , you can avoid computation by noticing that there is a denominator of
, so the answer must have a factor of
in the denominator, which only
does.
Solution 2
Let the square have side length . If we were to extend the sides of the square further into the triangle until they intersect on point on the hypotenuse, we'd have a similar right triangle formed between the hypotenuse and the two new lines, and 2 smaller similar triangles that share a side of length 2. Using the side-to-side ratios of these triangles, we can find that the length of the larger similar triangle is
. Now, let's extend this larger similar right triangle to the left until it hits the side of length 3. Now, the length is
, and using the ratios of the side lengths, the height is
. Looking at the diagram, if we add the height of this triangle to the side length of the square, we'd get 3, so
Now comes the easy part: finding the ratio of the areas: .
Solution 3
We use coordinate geometry. Let the right angle be at and the hypotenuse be the line
for
. Denote the position of
as
, and by the point to line distance formula, we know that
Solving this, we get
. Obviously
, so
, and from here, the rest of the solution follows to get
.
Solution 4
Let the side length of the square be . First off, let us make a similar triangle with the segment of length
and the top-right corner of
. Therefore, the longest side of the smaller triangle must be
. We then do operations with that side in terms of
. We subtract
from the bottom, and
from the top. That gives us the equation of
. Solving,
Thus, , so the fraction of the triangle (area
) covered by the square is
. The answer is then
.
Solution 5
On the diagram above, find two smaller triangles similar to the large one with side lengths ,
, and
; consequently, the segments with length
and
.
With being the side length of the square, we need to find an expression for
. Using the hypotenuse, we can see that
. Simplifying,
, or
.
A different calculation would yield , so
. In other words,
, while to check,
. As such,
, and
.
Finally, we get , to finish. As a proportion of the triangle with area
, the answer would be
, so
is correct.
Solution 6: Also Coordinate Geo
Let the right angle be at , the point
be the far edge of the unplanted square and the hypotenuse be the line
. Since the line from
to the hypotenuse is the shortest possible distance, we know this line, call it line
, is perpendicular to the hypotenuse and therefore has a slope of
.
Since we know , we can see that the line rises by
and moves to the right by
to meet the hypotenuse. (Let
and the rise be
and the run be
and then solve.) Therefore, line
intersects the hypotenuse at the point
. Plugging into the equation for the hypotenuse we have
, and after a bit of computation we get
Solution 7(Slightly different from first solution):
Same drawing as before:
Let's assign as the side length of box S. We then get each of the smaller triangle areas.
The sum of all the triangular areas(not including the box) is equal to
You can solve for
Then, the ratio would be which is equal to
~Starshooter11
Solution 8(Similar but cleaner than first solution):
Instead of dividing the large triangle into three right triangles plus a square, simply draw one diagonal of the square (can someone asymptote this for me I don't know how) divide the large triangle into three triangles with the sides and
as their bases. The total area is
as found above, and the square's side length is
. The area equation is:
which solves to
the same as the first solution (but easier to calculate). The final answer is
.
Hardness of Problem
The problem itself requires the drawing of a few obvious lines and algebra, although the image deceives the solver.
Comment: Hardness and difficulty are relative/subjective to the problem solver. It can be useful in personal situations to label questions with a hardness scale, but it may not suffice for other problem solvers. Please don't feel pressured to spontaneously know concepts that are difficult for you. Additionally, nothing is 'obvious' because it's all relative to the problem solver and individual minds. Happy mathing!
Video Solution by Richard Rusczyk
https://www.youtube.com/watch?v=p9npzq4FY_Y
~ dolphin7
See Also
2018 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 AMC 10 Problems and Solutions |
2018 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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 AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.