Difference between revisions of "2018 AMC 10A Problems/Problem 13"
Alexwin0806 (talk | contribs) m (→Solution 1) |
MRENTHUSIASM (talk | contribs) m |
||
(10 intermediate revisions by 8 users not shown) | |||
Line 1: | Line 1: | ||
+ | {{duplicate|[[2018 AMC 10A Problems/Problem 13|2018 AMC 10A #13]] and [[2018 AMC 12A Problems/Problem 11|2018 AMC 12A #11]]}} | ||
+ | |||
== Problem == | == Problem == | ||
− | + | A paper triangle with sides of lengths <math>3,4,</math> and <math>5</math> inches, as shown, is folded so that point <math>A</math> falls on point <math>B</math>. What is the length in inches of the crease? | |
− | A paper triangle with sides of lengths 3,4, and 5 inches, as shown, is folded so that point <math>A</math> falls on point <math>B</math>. What is the length in inches of the crease? | ||
<asy> | <asy> | ||
draw((0,0)--(4,0)--(4,3)--(0,0)); | draw((0,0)--(4,0)--(4,3)--(0,0)); | ||
Line 14: | Line 15: | ||
<math>\textbf{(A) } 1+\frac12 \sqrt2 \qquad \textbf{(B) } \sqrt3 \qquad \textbf{(C) } \frac74 \qquad \textbf{(D) } \frac{15}{8} \qquad \textbf{(E) } 2 </math> | <math>\textbf{(A) } 1+\frac12 \sqrt2 \qquad \textbf{(B) } \sqrt3 \qquad \textbf{(C) } \frac74 \qquad \textbf{(D) } \frac{15}{8} \qquad \textbf{(E) } 2 </math> | ||
− | ==Solution 1== | + | == Solution 1 == |
− | <asy> draw((0,0)--(4,0)--(4,3)--(0,0)); label("$A$", (0,0), SW); label("$B$", (4,3), NE); label("$C$", (4,0), SE); label("$D$", (2,1.5), NW); label("$E$", (3.125,0), S); | + | <asy> |
+ | draw((0,0)--(4,0)--(4,3)--(0,0)); label("$A$", (0,0), SW); label("$B$", (4,3), NE); label("$C$", (4,0), SE); label("$D$", (2,1.5), NW); label("$E$", (3.125,0), S); | ||
draw ((2,1.5)--(3.125,0),linewidth(1)); | draw ((2,1.5)--(3.125,0),linewidth(1)); | ||
− | draw(rightanglemark((0,0),(2,1.5),(3.125,0))); </asy> | + | draw(rightanglemark((0,0),(2,1.5),(3.125,0))); |
− | + | </asy> | |
First, we need to realize that the crease line is just the perpendicular bisector of side <math>AB</math>, the hypotenuse of right triangle <math>\triangle ABC</math>. Call the midpoint of <math>AB</math> point <math>D</math>. Draw this line and call the intersection point with <math>AC</math> as <math>E</math>. Now, <math>\triangle ACB</math> is similar to <math>\triangle ADE</math> by <math>AA</math> similarity. Setting up the ratios, we find that | First, we need to realize that the crease line is just the perpendicular bisector of side <math>AB</math>, the hypotenuse of right triangle <math>\triangle ABC</math>. Call the midpoint of <math>AB</math> point <math>D</math>. Draw this line and call the intersection point with <math>AC</math> as <math>E</math>. Now, <math>\triangle ACB</math> is similar to <math>\triangle ADE</math> by <math>AA</math> similarity. Setting up the ratios, we find that | ||
Line 24: | Line 26: | ||
Thus, our answer is <math>\boxed{\textbf{(D) } \frac{15}{8}}</math>. | Thus, our answer is <math>\boxed{\textbf{(D) } \frac{15}{8}}</math>. | ||
− | + | ==== Note ==== | |
− | ===Note=== | ||
In general, whenever we are asked to make a crease, think about that crease as a line of reflection over which the diagram is reflected. This is why the crease must be the perpendicular bisector of <math>AB</math>, because <math>A</math> must be reflected onto <math>B</math>. | In general, whenever we are asked to make a crease, think about that crease as a line of reflection over which the diagram is reflected. This is why the crease must be the perpendicular bisector of <math>AB</math>, because <math>A</math> must be reflected onto <math>B</math>. | ||
− | ==Solution 2 (Draw it, not recommended)== | + | == Solution 2 (Draw it, not recommended) == |
− | + | Use the ruler and graph paper you brought to quickly draw a 3-4-5 triangle of any scale (don't trust the diagram in the booklet). Very carefully fold the acute vertices together and make a crease. Measure the crease with the ruler. If you were reasonably careful, you should see that it measures somewhat more than <math>\frac{7}{4}</math> units and somewhat less than <math>2</math> units. The only answer choice in that range is <math>\boxed{\textbf{D) } \frac{15}{8}}</math>. | |
− | Use the ruler and graph paper you brought to quickly draw a 3-4-5 triangle of any scale (don't trust the diagram in the booklet). Very carefully fold the acute vertices together and make a crease. Measure the crease with the ruler. If you were reasonably careful, you should see that it measures somewhat more than <math>\frac{7}{4}</math> units and somewhat less than <math>2</math> units. The only answer choice in range is <math>\boxed{\textbf{D) } \frac{15}{8}}</math>. | ||
This is pretty much a cop-out, but it's allowed in the rules technically. This is basically useless for problems without diagrams. | This is pretty much a cop-out, but it's allowed in the rules technically. This is basically useless for problems without diagrams. | ||
− | ==Solution 3== | + | == Solution 3 == |
− | |||
Since <math>\triangle ABC</math> is a right triangle, we can see that the slope of line <math>AB</math> is <math>\frac{BC}{AC}</math> = <math>\frac{3}{4}</math>. We know that if we fold <math>\triangle ABC</math> so that point <math>A</math> meets point <math>B</math> the crease line will be perpendicular to <math>AB</math> and we also know that the slopes of perpendicular lines are negative reciprocals of one another. Then, we can see that the slope of our crease line is <math>-\frac{4}{3}</math>. | Since <math>\triangle ABC</math> is a right triangle, we can see that the slope of line <math>AB</math> is <math>\frac{BC}{AC}</math> = <math>\frac{3}{4}</math>. We know that if we fold <math>\triangle ABC</math> so that point <math>A</math> meets point <math>B</math> the crease line will be perpendicular to <math>AB</math> and we also know that the slopes of perpendicular lines are negative reciprocals of one another. Then, we can see that the slope of our crease line is <math>-\frac{4}{3}</math>. | ||
<asy> | <asy> | ||
Line 60: | Line 59: | ||
label("$5$", (2,1.5), NW); | label("$5$", (2,1.5), NW); | ||
</asy> | </asy> | ||
+ | |||
Let us call the midpoint of <math>AB</math> point <math>D</math>, the midpoint of <math>AC</math> point <math>E</math>, and the crease line <math>DF</math>. We know that <math>DE</math> is parallel to <math>BC</math> and that <math>DE</math>'s length is <math>\frac{BC}{2}=\frac{3}{2}</math>. Using our slope calculation from earlier, we can see that<math>-\frac{DE}{EF}=-\frac{\frac{3}{2}}{EF}=-\frac{4}{3}</math>. With this information, we can solve for <math>EF</math>: | Let us call the midpoint of <math>AB</math> point <math>D</math>, the midpoint of <math>AC</math> point <math>E</math>, and the crease line <math>DF</math>. We know that <math>DE</math> is parallel to <math>BC</math> and that <math>DE</math>'s length is <math>\frac{BC}{2}=\frac{3}{2}</math>. Using our slope calculation from earlier, we can see that<math>-\frac{DE}{EF}=-\frac{\frac{3}{2}}{EF}=-\frac{4}{3}</math>. With this information, we can solve for <math>EF</math>: | ||
<cmath>-4EF=(-\frac{3}{2})(3) \Rightarrow -4EF=-\frac{9}{2} \Rightarrow 4EF=\frac{9}{2} \Rightarrow EF=\frac{9}{8}.</cmath> | <cmath>-4EF=(-\frac{3}{2})(3) \Rightarrow -4EF=-\frac{9}{2} \Rightarrow 4EF=\frac{9}{2} \Rightarrow EF=\frac{9}{8}.</cmath> | ||
Line 67: | Line 67: | ||
Thus, our answer is <math>\boxed{\textbf{D) } \frac{15}{8}}</math>. | Thus, our answer is <math>\boxed{\textbf{D) } \frac{15}{8}}</math>. | ||
+ | == Solution 4 == | ||
− | |||
Make use of the diagram in Solution 3. It can be deduced that <math>AF=BF</math>. Let <math>DF=x</math>. In <math>\triangle ADF</math>, <math>AF^2=x^2+2.5^2 \Rightarrow AF=\sqrt{x^2+2.5^2}</math>. Then <math>FC</math> also would be <math>4-\sqrt{x^2+2.5^2}</math>. | Make use of the diagram in Solution 3. It can be deduced that <math>AF=BF</math>. Let <math>DF=x</math>. In <math>\triangle ADF</math>, <math>AF^2=x^2+2.5^2 \Rightarrow AF=\sqrt{x^2+2.5^2}</math>. Then <math>FC</math> also would be <math>4-\sqrt{x^2+2.5^2}</math>. | ||
Line 76: | Line 76: | ||
<math>\therefore</math> <math>DF=x=</math> <math>\boxed{\textbf{(D) } \frac{15}{8}}</math>. | <math>\therefore</math> <math>DF=x=</math> <math>\boxed{\textbf{(D) } \frac{15}{8}}</math>. | ||
+ | == Solution 5 == | ||
+ | Like in Solution 3, we can make use of coordinate geometry to solve this problem. Because the length of the crease is constant no matter the positioning of the triangle, reorient the triangle so it has C at the origin and B and A at <math>(0,3)</math> and <math>(4,0)</math> respectively. | ||
− | |||
− | |||
− | |||
<asy> | <asy> | ||
draw((0,0)--(0,3)--(4,0)--(0,0)); | draw((0,0)--(0,3)--(4,0)--(0,0)); | ||
Line 98: | Line 97: | ||
<math>(2-\frac{7}{8})^2+({\frac{3}{2}})^2=d^2 \implies d^2=\frac{225}{64} \implies d=\boxed{\frac{15}{8}}</math> | <math>(2-\frac{7}{8})^2+({\frac{3}{2}})^2=d^2 \implies d^2=\frac{225}{64} \implies d=\boxed{\frac{15}{8}}</math> | ||
− | + | == Solution 6 == | |
− | ==Solution 6== | ||
Let <math>D</math> be a point on AC such that <math>AD=DB</math> and <math>\triangle{ADB}</math> is isosceles. The altitude of this triangle is the crease of the paper triangle. | Let <math>D</math> be a point on AC such that <math>AD=DB</math> and <math>\triangle{ADB}</math> is isosceles. The altitude of this triangle is the crease of the paper triangle. | ||
Line 111: | Line 109: | ||
Our answer is <math>D</math>. | Our answer is <math>D</math>. | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/4_x1sgcQCp4?t=2623 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
== See Also == | == See Also == |
Latest revision as of 02:46, 5 January 2022
- The following problem is from both the 2018 AMC 10A #13 and 2018 AMC 12A #11, so both problems redirect to this page.
Contents
Problem
A paper triangle with sides of lengths and inches, as shown, is folded so that point falls on point . What is the length in inches of the crease?
Solution 1
First, we need to realize that the crease line is just the perpendicular bisector of side , the hypotenuse of right triangle . Call the midpoint of point . Draw this line and call the intersection point with as . Now, is similar to by similarity. Setting up the ratios, we find that Thus, our answer is .
Note
In general, whenever we are asked to make a crease, think about that crease as a line of reflection over which the diagram is reflected. This is why the crease must be the perpendicular bisector of , because must be reflected onto .
Solution 2 (Draw it, not recommended)
Use the ruler and graph paper you brought to quickly draw a 3-4-5 triangle of any scale (don't trust the diagram in the booklet). Very carefully fold the acute vertices together and make a crease. Measure the crease with the ruler. If you were reasonably careful, you should see that it measures somewhat more than units and somewhat less than units. The only answer choice in that range is .
This is pretty much a cop-out, but it's allowed in the rules technically. This is basically useless for problems without diagrams.
Solution 3
Since is a right triangle, we can see that the slope of line is = . We know that if we fold so that point meets point the crease line will be perpendicular to and we also know that the slopes of perpendicular lines are negative reciprocals of one another. Then, we can see that the slope of our crease line is .
Let us call the midpoint of point , the midpoint of point , and the crease line . We know that is parallel to and that 's length is . Using our slope calculation from earlier, we can see that. With this information, we can solve for : We can then use the Pythagorean Theorem to find . Thus, our answer is .
Solution 4
Make use of the diagram in Solution 3. It can be deduced that . Let . In , . Then also would be .
In , . After some quick math, we get . Solving for will give .
.
Solution 5
Like in Solution 3, we can make use of coordinate geometry to solve this problem. Because the length of the crease is constant no matter the positioning of the triangle, reorient the triangle so it has C at the origin and B and A at and respectively.
We know that each point in the crease is equidistant from and , so the crease must pass through the midpoint of , which is , and be perpendicular to hypotenuse . The crease therefore has a slope of .
Plugging into point-slope we find that the equation of the crease is,
Using 0 for y, we see that the crease intersects at .
By the Distance Formula,
Solution 6
Let be a point on AC such that and is isosceles. The altitude of this triangle is the crease of the paper triangle.
Setting , we have a right triangle with legs of x and 3, and, because hypotenuse of .
The area of Because we can plug in and see that
Since , we get
Our answer is .
Video Solution
https://youtu.be/4_x1sgcQCp4?t=2623
~ pi_is_3.14
See Also
2018 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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 10 |
Followed by Problem 12 |
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.