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2018 AMC 10A Problems/Problem 13

Revision as of 08:48, 24 January 2019 by Olutosinfires (talk | contribs) (→‎Solution 4)

Problem

A paper triangle with sides of lengths 3,4, and 5 inches, as shown, is folded so that point $A$ falls on point $B$. What is the length in inches of the crease? [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("$4$", (2,0), S); label("$3$", (4,1.5), E); label("$5$", (2,1.5), NW); fill(origin--(0,0)--(4,3)--(4,0)--cycle, gray); [/asy] $\textbf{(A) } 1+\frac12 \sqrt2 \qquad \textbf{(B) } \sqrt3 \qquad \textbf{(C) } \frac74 \qquad \textbf{(D) } \frac{15}{8} \qquad \textbf{(E) } 2$

Solution 1

First, we need to realize that the crease line is just the perpendicular bisector of side $AB$, the hypotenuse of right triangle $\triangle ABC$. Call the midpoint of $AB$ point $D$. Draw this line and call the intersection point with $AC$ as $E$. Now, $\triangle ACB$ is similar to $\triangle ADE$ by $AA$ similarity. Setting up the ratios, we find that \[\frac{BC}{AC}=\frac{DE}{AD} \Rightarrow \frac{3}{4}=\frac{DE}{\frac{5}{2}} \Rightarrow DE=\frac{15}{8}.\] Thus, our answer is $\boxed{\textbf{D) } \frac{15}{8}}$.

~Nivek

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 $AB$, because $A$ must be reflected onto $B$. (by pulusona)

Solution 2

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 $\frac{7}{4}$ units and somewhat less than $2$ units. The only answer choice in range is $\boxed{\textbf{D) } \frac{15}{8}}$.

This is pretty much a cop-out, but it's allowed in the rules technically.

Solution 3

Since $\triangle ABC$ is a right triangle, we can see that the slope of line $AB$ is $\frac{BC}{AC}$ = $\frac{3}{4}$. We know that if we fold $\triangle ABC$ so that point $A$ meets point $B$ the crease line will be perpendicular to $AB$ 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 $-\frac{4}{3}$. [asy] pen dotstyle = black; draw((0,0)--(4,0)--(4,3)--(0,0)); fill(origin--(0,0)--(4,3)--(4,0)--cycle, gray); dot((0,0),dotstyle); label("$A$", (0.03153837092244126,0.07822624343603715), SW); dot((4,0),dotstyle); label("$C$", (4.028913881471271,0.07822624343603715), SE); dot((4,3),dotstyle); label("$B$", (4.028913881471271,3.078221223847919), NE); dot((2,1.5),dotstyle); label("$D$", (2.0341528211973956,1.578223733641978), SE); dot((2,0),dotstyle); label("$E$", (2.0341528211973956,0.07822624343603715), NE); dot((3.1249518689638927,0),dotstyle); label("$F$", (3.1571875913515854,0.07822624343603715), NE); label("$4$", (2,0), S); label("$3$", (4,1.5), E); label("$5$", (2,1.5), NW); [/asy] Let us call the midpoint of $AB$ point $D$, the midpoint of $AC$ point $E$, and the crease line $DF$. We know that $DE$ is parallel to $AC$ and that $DE$'s length is $\frac{AC}{2}=\frac{3}{2}$. Using our slope calculation from earlier, we can see that$-\frac{DE}{EF}=-\frac{\frac{3}{2}}{EF}=-\frac{4}{3}$. With this information, we can solve for $EF$: \[-4EF=(-\frac{3}{2})(3) \Rightarrow -4EF=-\frac{9}{2} \Rightarrow 4EF=\frac{9}{2} \Rightarrow EF=\frac{9}{8}.\] We can then use the Pythagorean Theorem to find $DF$. \[\frac{3}{2}^2+\frac{9}{8}^2=DF^2 \Rightarrow \frac{9}{4}+\frac{9\cdot9}{8\cdot8}=DF^2 \Rightarrow \frac{9\cdot2\cdot8}{4\cdot2\cdot8}+\frac{9\cdot9}{8\cdot8}=DF^2 \Rightarrow \frac{9\cdot2\cdot8+9\cdot9}{8\cdot8}=DF^2 \Rightarrow \frac{9(2\cdot8+9)}{8\cdot8}=DF^2\] \[\Rightarrow DF=\sqrt{\frac{9(2\cdot8+9)}{8\cdot8}} \Rightarrow DF=\frac{3\cdot5}{8} \Rightarrow DF=\frac{15}{8}\] Thus, our answer is $\boxed{\textbf{D) } \frac{15}{8}}$.


$\propto M o o n R a b b i t$


Solution 4

Make use of the diagram in Solution 3. It can be deduced that $AF=BF$. Let $DF=x$. In $\triangle ADF$, $AF^2=x^2+2.5^2 \Rightarrow AF=\sqrt{x^2+2.5^2}$. Then $FC$ also would be $4-\sqrt{x^2+2.5^2}$.

In $\triangle BCF$, $BF^2=FC^2+BC^2 \Rightarrow (\sqrt{x^2+2.5^2})^2=(4-\sqrt{x^2+2.5^2})^2+3^2$. After some quick math, we get $\sqrt{x^2+2.5^2}=\frac{25}{8}$. Solving for $x$ will give $x=\frac{15}{8}$.

$\therefore$ $DF=x=$ $\boxed{\textbf{(D) } \frac{15}{8}}$.

~OlutosinNGA

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

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