Difference between revisions of "2018 AMC 10A Problems/Problem 23"
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<math>\textbf{(A) } \frac{25}{27} \qquad \textbf{(B) } \frac{26}{27} \qquad \textbf{(C) } \frac{73}{75} \qquad \textbf{(D) } \frac{145}{147} \qquad \textbf{(E) } \frac{74}{75} </math> | <math>\textbf{(A) } \frac{25}{27} \qquad \textbf{(B) } \frac{26}{27} \qquad \textbf{(C) } \frac{73}{75} \qquad \textbf{(D) } \frac{145}{147} \qquad \textbf{(E) } \frac{74}{75} </math> | ||
− | ==Solution 1== | + | ==Solution 1 (Area Addition)== |
− | + | Note that the hypotenuse of the field is <math>5,</math> and the area of the field is <math>6.</math> Let <math>x</math> be the side-length of square <math>S.</math> | |
We partition the field into a red triangle, a yellow triangle, and a green triangle, as shown below: | We partition the field into a red triangle, a yellow triangle, and a green triangle, as shown below: | ||
Line 28: | Line 28: | ||
draw((0,0)--(4,0)--(0,3)--(0,0)); | draw((0,0)--(4,0)--(0,3)--(0,0)); | ||
draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0)); | draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0)); | ||
+ | label("$5$", (2,1.5), NE); | ||
label("$4$", (2,0), S); | label("$4$", (2,0), S); | ||
label("$3$", (0,1.5), W); | label("$3$", (0,1.5), W); | ||
Line 39: | Line 40: | ||
label("$\small{x}$", (0.3,0.15), E); | label("$\small{x}$", (0.3,0.15), E); | ||
</asy> | </asy> | ||
+ | Let the brackets denote areas. By area addition, we set up an equation for <math>x:</math> | ||
+ | <cmath>\begin{align*} | ||
+ | [\text{Red Triangle}]+[\text{Yellow Triangle}]+[\text{Green Triangle}]&=[\text{Field}] \\ | ||
+ | \frac{3x}{2}+\frac{4x}{2}+\frac{5\cdot2}{2}&=6 \\ | ||
+ | \frac{7x}{2}+5&=6, | ||
+ | \end{align*}</cmath> | ||
+ | from which <math>x=\frac27.</math> Therefore, the answer is <cmath>\frac{[\text{Field}]-[S]}{[\text{Field}]}=\frac{6-x^2}{6}=\boxed{\textbf{(D) } \frac{145}{147}}.</cmath> | ||
+ | ~MRENTHUSIASM | ||
− | ==Solution 2== | + | ==Solution 2 (Area Addition)== |
Let the square have side length <math>x</math>. Connect the upper-right vertex of square <math>S</math> 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 <math>6</math>. | Let the square have side length <math>x</math>. Connect the upper-right vertex of square <math>S</math> 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 <math>6</math>. | ||
− | |||
<asy> | <asy> | ||
draw((0,0)--(4,0)--(0,3)--(0,0)); | draw((0,0)--(4,0)--(0,3)--(0,0)); | ||
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label("$\small{x}$", (0.3,0.15), E); | label("$\small{x}$", (0.3,0.15), E); | ||
</asy> | </asy> | ||
− | + | Square <math>S</math> has area <math>x^2</math>, and the two thin triangle regions have area <math>\dfrac{x(3-x)}{2}</math> and <math>\dfrac{x(4-x)}{2}</math>. The final triangular region with the hypotenuse as its base and height <math>2</math> has area <math>5</math>. Thus, we have <cmath>x^2+\dfrac{x(3-x)}{2}+\dfrac{x(4-x)}{2}+5=6.</cmath> | |
− | Square <math>S</math> has area <math>x^2</math>, and the two thin triangle regions have area <math>\dfrac{x(3-x)}{2}</math> and <math>\dfrac{x(4-x)}{2}</math>. The final triangular region with the hypotenuse as its base and height <math>2</math> has area <math>5</math>. Thus, we have <cmath>x^2+\dfrac{x(3-x)}{2}+\dfrac{x(4-x)}{2}+5=6</cmath> | ||
− | |||
Solving gives <math>x=\dfrac{2}{7}</math>. The area of <math>S</math> is <math>\dfrac{4}{49}</math> and the desired ratio is <math>\dfrac{6-\dfrac{4}{49}}{6}=\boxed{\textbf{(D) } \frac{145}{147}}</math>. | Solving gives <math>x=\dfrac{2}{7}</math>. The area of <math>S</math> is <math>\dfrac{4}{49}</math> and the desired ratio is <math>\dfrac{6-\dfrac{4}{49}}{6}=\boxed{\textbf{(D) } \frac{145}{147}}</math>. | ||
Alternatively, once you get <math>x=\frac{2}{7}</math>, you can avoid computation by noticing that there is a denominator of <math>7</math>, so the answer must have a factor of <math>7</math> in the denominator, which only <math>\frac{145}{147}</math> does. | Alternatively, once you get <math>x=\frac{2}{7}</math>, you can avoid computation by noticing that there is a denominator of <math>7</math>, so the answer must have a factor of <math>7</math> in the denominator, which only <math>\frac{145}{147}</math> does. | ||
− | ==Solution 3== | + | ==Solution 3 (Similar Triangles)== |
Let the square have side length <math>s</math>. 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 <math>2</math> smaller similar triangles that share a side of length <math>2</math>. Using the side-to-side ratios of these triangles, we can find that the length of the larger similar triangle is <math>\frac{5}{3}(2)=\frac{10}{3}</math>. Now, let's extend this larger similar right triangle to the left until it hits the side of length <math>3</math>. Now, the length is <math>\frac{10}{3}+s</math>, and using the ratios of the side lengths, the height is <math>\frac{3}{4}\left(\frac{10}{3}+s\right)=\frac{5}{2}+\frac{3s}{4}</math>. Looking at the diagram, if we add the height of this triangle to the side length of the square, we'd get <math>3</math>, so <cmath>\frac{5}{2}+\frac{3s}{4}+s=\frac{5}{2}+\frac{7s}{4}=3 \\ \frac{7s}{4}=\frac{1}{2} \\ s=\frac{2}{7} \implies \textrm{ area of square is } \left(\frac{2}{7}\right)^2=\frac{4}{49}.</cmath> | Let the square have side length <math>s</math>. 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 <math>2</math> smaller similar triangles that share a side of length <math>2</math>. Using the side-to-side ratios of these triangles, we can find that the length of the larger similar triangle is <math>\frac{5}{3}(2)=\frac{10}{3}</math>. Now, let's extend this larger similar right triangle to the left until it hits the side of length <math>3</math>. Now, the length is <math>\frac{10}{3}+s</math>, and using the ratios of the side lengths, the height is <math>\frac{3}{4}\left(\frac{10}{3}+s\right)=\frac{5}{2}+\frac{3s}{4}</math>. Looking at the diagram, if we add the height of this triangle to the side length of the square, we'd get <math>3</math>, so <cmath>\frac{5}{2}+\frac{3s}{4}+s=\frac{5}{2}+\frac{7s}{4}=3 \\ \frac{7s}{4}=\frac{1}{2} \\ s=\frac{2}{7} \implies \textrm{ area of square is } \left(\frac{2}{7}\right)^2=\frac{4}{49}.</cmath> | ||
Now comes the easy part: finding the ratio of the areas: <math>\frac{3\cdot 4 \cdot \frac{1}{2} -\frac{4}{49}}{3\cdot 4 \cdot \frac{1}{2}}=\frac{6-\frac{4}{49}}{6}=\frac{294-4}{294}=\frac{290}{294}=\boxed{\textbf{(D) } \frac{145}{147}}</math>. | Now comes the easy part: finding the ratio of the areas: <math>\frac{3\cdot 4 \cdot \frac{1}{2} -\frac{4}{49}}{3\cdot 4 \cdot \frac{1}{2}}=\frac{6-\frac{4}{49}}{6}=\frac{294-4}{294}=\frac{290}{294}=\boxed{\textbf{(D) } \frac{145}{147}}</math>. | ||
− | ==Solution 4== | + | ==Solution 4 (Similar Triangles)== |
<asy> | <asy> | ||
draw((0,0)--(4,0)--(0,3)--(0,0)); | draw((0,0)--(4,0)--(0,3)--(0,0)); | ||
Line 91: | Line 97: | ||
label("$\small{l}$", (0.15,0.3), N); | label("$\small{l}$", (0.15,0.3), N); | ||
</asy> | </asy> | ||
− | |||
On the diagram above, find two smaller triangles similar to the large one with side lengths <math>3</math>, <math>4</math>, and <math>5</math>; consequently, the segments with length <math>\frac{5}{2}</math> and <math>\frac{10}{3}</math>. | On the diagram above, find two smaller triangles similar to the large one with side lengths <math>3</math>, <math>4</math>, and <math>5</math>; consequently, the segments with length <math>\frac{5}{2}</math> and <math>\frac{10}{3}</math>. | ||
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Finally, we get <math>A(\Square S)=l^2=\frac{4}{49}</math>, to finish. As a proportion of the triangle with area <math>6</math>, the answer would be <math>1-\frac{4}{49\cdot6}=1-\frac{2}{147}=\frac{145}{147}</math>, so <math>\boxed{\textbf{(D) } \frac{145}{147}}</math> is correct. | Finally, we get <math>A(\Square S)=l^2=\frac{4}{49}</math>, to finish. As a proportion of the triangle with area <math>6</math>, the answer would be <math>1-\frac{4}{49\cdot6}=1-\frac{2}{147}=\frac{145}{147}</math>, so <math>\boxed{\textbf{(D) } \frac{145}{147}}</math> is correct. | ||
− | ==Solution 5== | + | ==Solution 5 (Similar Triangles)== |
Let the side length of the square be <math>x</math>. First off, let us make a similar triangle with the segment of length <math>2</math> and the top-right corner of <math>S</math>. Therefore, the longest side of the smaller triangle must be <math>2 \cdot \frac54 = \frac52</math>. We then do operations with that side in terms of <math>x</math>. We subtract <math>x</math> from the bottom, and <math>\frac{3x}{4}</math> from the top. That gives us the equation of <math>3-\frac{7x}{4} = \frac{5}{2}</math>. Solving, <cmath>12-7x = 10 \implies x = \frac{2}{7}.</cmath> | Let the side length of the square be <math>x</math>. First off, let us make a similar triangle with the segment of length <math>2</math> and the top-right corner of <math>S</math>. Therefore, the longest side of the smaller triangle must be <math>2 \cdot \frac54 = \frac52</math>. We then do operations with that side in terms of <math>x</math>. We subtract <math>x</math> from the bottom, and <math>\frac{3x}{4}</math> from the top. That gives us the equation of <math>3-\frac{7x}{4} = \frac{5}{2}</math>. Solving, <cmath>12-7x = 10 \implies x = \frac{2}{7}.</cmath> | ||
Line 106: | Line 111: | ||
==Solution 6 (Coordinate Geometry)== | ==Solution 6 (Coordinate Geometry)== | ||
− | We use coordinate geometry. Let the right angle be at <math>(0,0)</math> and the hypotenuse be the line <math>3x+4y = 12</math> for <math>0\le x\le 3</math>. Denote the position of <math>S</math> as <math>(s,s)</math>, and by the point to line distance formula, we know that <cmath>\frac{|3s+4s-12|}{5} = 2 | + | We use coordinate geometry. Let the right angle be at <math>(0,0)</math> and the hypotenuse be the line <math>3x+4y = 12</math> for <math>0\le x\le 3</math>. Denote the position of <math>S</math> as <math>(s,s)</math>, and by the point to line distance formula, we know that |
+ | <cmath>\begin{align*} | ||
+ | \frac{|3s+4s-12|}{5} &= 2 \\ | ||
+ | |7s-12| &= 10 | ||
+ | \end{align*}</cmath> | ||
+ | Solving this, we get <math>s=\frac{22}{7}, \frac{2}{7}</math>. Obviously <math>s<\frac{22}{7}</math>, so <math>s = \frac{2}{7}</math>, and from here, the rest of the solution follows to get <math>\boxed{\textbf{(D) } \frac{145}{147}}</math>. | ||
==Solution 7 (Coordinate Geometry)== | ==Solution 7 (Coordinate Geometry)== | ||
Let the right angle be at <math>(0,0)</math>, the point <math>(x,x)</math> be the far edge of the unplanted square and the hypotenuse be the line <math>y=-\frac{3}{4}x+3</math>. Since the line from <math>(x,x)</math> to the hypotenuse is the shortest possible distance, we know this line, call it line <math>\l</math>, is perpendicular to the hypotenuse and therefore has a slope of <math>\frac{4}{3}</math>. | Let the right angle be at <math>(0,0)</math>, the point <math>(x,x)</math> be the far edge of the unplanted square and the hypotenuse be the line <math>y=-\frac{3}{4}x+3</math>. Since the line from <math>(x,x)</math> to the hypotenuse is the shortest possible distance, we know this line, call it line <math>\l</math>, is perpendicular to the hypotenuse and therefore has a slope of <math>\frac{4}{3}</math>. | ||
− | Since we know <math>m=\frac{4}{3}</math> , we can see that the line rises by <math>\frac{8}{5}</math> and moves to the right by <math>\frac{6}{5}</math> to meet the hypotenuse. (Let <math>2 = 5x</math> and the rise be <math>4x</math> and the run be <math>3x</math> and then solve.) Therefore, line <math>\l</math> intersects the hypotenuse at the point <math>(x+\frac{6}{5}, x+\frac{8}{5})</math>. Plugging into the equation for the hypotenuse we have <math>x=\frac{2}{7}</math> , and after a bit of computation we get <math>\boxed{\textbf{(D) } \frac{145}{147}}</math>. | + | Since we know <math>m=\frac{4}{3}</math> , we can see that the line rises by <math>\frac{8}{5}</math> and moves to the right by <math>\frac{6}{5}</math> to meet the hypotenuse. (Let <math>2 = 5x</math> and the rise be <math>4x</math> and the run be <math>3x</math> and then solve.) Therefore, line <math>\l</math> intersects the hypotenuse at the point <math>\left(x+\frac{6}{5}, x+\frac{8}{5}\right)</math>. Plugging into the equation for the hypotenuse we have <math>x=\frac{2}{7}</math> , and after a bit of computation we get <math>\boxed{\textbf{(D) } \frac{145}{147}}</math>. |
==Hardness of Problem== | ==Hardness of Problem== |
Revision as of 15:10, 17 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
- 1 Problem
- 2 Solution 1 (Area Addition)
- 3 Solution 2 (Area Addition)
- 4 Solution 3 (Similar Triangles)
- 5 Solution 4 (Similar Triangles)
- 6 Solution 5 (Similar Triangles)
- 7 Solution 6 (Coordinate Geometry)
- 8 Solution 7 (Coordinate Geometry)
- 9 Hardness of Problem
- 10 Video Solution by Richard Rusczyk
- 11 See Also
Problem
Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths and
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
units. What fraction of the field is planted?
Solution 1 (Area Addition)
Note that the hypotenuse of the field is and the area of the field is
Let
be the side-length of square
We partition the field into a red triangle, a yellow triangle, and a green triangle, as shown below:
Let the brackets denote areas. By area addition, we set up an equation for
from which
Therefore, the answer is
~MRENTHUSIASM
Solution 2 (Area Addition)
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 3 (Similar Triangles)
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
smaller similar triangles that share a side of length
. 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
. 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
, so
Now comes the easy part: finding the ratio of the areas:
.
Solution 4 (Similar Triangles)
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 5 (Similar Triangles)
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 6 (Coordinate Geometry)
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 7 (Coordinate Geometry)
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
.
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.