Difference between revisions of "2018 AMC 10A Problems/Problem 13"

m (Solution)
(Solution)
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<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==
+
==Solution 1==
 
 
Solution 1:
 
  
 
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>AC</math> point <math>D</math>. Draw this line and call the intersection point with <math>AC</math> as <math>E</math>. Now, <math>\triangle ABC</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>AC</math> point <math>D</math>. Draw this line and call the intersection point with <math>AC</math> as <math>E</math>. Now, <math>\triangle ABC</math> is similar to <math>\triangle ADE</math> by <math>AA</math> similarity. Setting up the ratios, we find that
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~Nivek
 
~Nivek
  
Solution 2 (if you are already out of time):
+
==Solution 2 (if you are already out of time)==
  
 
Simply make a 3x4x5 inch triangle and then cut it out (using fine rips). Then, make the fold and mesure. It will be <math>\boxed{D} \frac{15}{8}</math> inches in length.
 
Simply make a 3x4x5 inch triangle and then cut it out (using fine rips). Then, make the fold and mesure. It will be <math>\boxed{D} \frac{15}{8}</math> inches in length.

Revision as of 16:59, 8 February 2018

A paper triangle with sides of lengths 3,4, and 5 inches, as shon, 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 $AC$ point $D$. Draw this line and call the intersection point with $AC$ as $E$. Now, $\triangle ABC$ 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 $\boxed{D}$.

~Nivek

Solution 2 (if you are already out of time)

Simply make a 3x4x5 inch triangle and then cut it out (using fine rips). Then, make the fold and mesure. It will be $\boxed{D} \frac{15}{8}$ inches in length.

See Also

2018 AMC 10A (ProblemsAnswer KeyResources)
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

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