Difference between revisions of "2000 AMC 10 Problems/Problem 16"
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(Buried at the end of Solution 2 is the "intended" solution. Break this out into its own Solution and explain it a little better.) |
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</asy> | </asy> | ||
− | <math>\ | + | <math>\textbf{(A)}\ \frac{4\sqrt{5}}{3} \qquad\textbf{(B)}\ \frac{5\sqrt{5}}{3} \qquad\textbf{(C)}\ \frac{12\sqrt{5}}{7} \qquad\textbf{(D)}\ 2\sqrt{5} \qquad\textbf{(E)}\ \frac{5\sqrt{65}}{9}</math> |
− | |||
− | ===Solution 1 | + | ==Video Solution== |
+ | https://youtu.be/oWxqYyW926I | ||
+ | |||
+ | ==Solution 1== | ||
Let <math>l_1</math> be the line containing <math>A</math> and <math>B</math> and let <math>l_2</math> be the line containing <math>C</math> and <math>D</math>. If we set the bottom left point at <math>(0,0)</math>, then <math>A=(0,3)</math>, <math>B=(6,0)</math>, <math>C=(4,2)</math>, and <math>D=(2,0)</math>. | Let <math>l_1</math> be the line containing <math>A</math> and <math>B</math> and let <math>l_2</math> be the line containing <math>C</math> and <math>D</math>. If we set the bottom left point at <math>(0,0)</math>, then <math>A=(0,3)</math>, <math>B=(6,0)</math>, <math>C=(4,2)</math>, and <math>D=(2,0)</math>. | ||
− | The line <math>l_1</math> is given by the equation <math>y=m_1x+b_1</math>. The <math>y</math>-intercept is <math>A=(0,3)</math>, so <math>b_1=3</math>. We are given two points on <math>l_1</math>, hence we can compute the slope, <math>m_1</math> to be <math>\frac{0-3}{6-0}=\frac{ | + | The line <math>l_1</math> is given by the equation <math>y=m_1x+b_1</math>. The <math>y</math>-intercept is <math>A=(0,3)</math>, so <math>b_1=3</math>. We are given two points on <math>l_1</math>, hence we can compute the slope, <math>m_1</math> to be <math>\frac{0-3}{6-0}=-\frac{1}{2}</math>, so <math>l_1</math> is the line <math>y=\frac{-1}{2}x+3</math> |
Similarly, <math>l_2</math> is given by <math>y=m_2x+b_2</math>. The slope in this case is <math>\frac{2-0}{4-2}=1</math>, so <math>y=x+b_2</math>. Plugging in the point <math>(2,0)</math> gives us <math>b_2=-2</math>, so <math>l_2</math> is the line <math>y=x-2</math>. | Similarly, <math>l_2</math> is given by <math>y=m_2x+b_2</math>. The slope in this case is <math>\frac{2-0}{4-2}=1</math>, so <math>y=x+b_2</math>. Plugging in the point <math>(2,0)</math> gives us <math>b_2=-2</math>, so <math>l_2</math> is the line <math>y=x-2</math>. | ||
Line 69: | Line 71: | ||
which is answer choice <math>\boxed{\text{B}}</math> | which is answer choice <math>\boxed{\text{B}}</math> | ||
− | + | ==Solution 2== | |
<asy> | <asy> | ||
Line 121: | Line 123: | ||
<math>m\angle AFE=m\angle BCE=90^\circ</math>, and <math>m\angle AEF=m\angle CEB</math>, so by AA similarity, <cmath>\triangle AFE\sim \triangle BCE \Rightarrow \frac{AF}{AE}=\frac{BC}{BE}</cmath> | <math>m\angle AFE=m\angle BCE=90^\circ</math>, and <math>m\angle AEF=m\angle CEB</math>, so by AA similarity, <cmath>\triangle AFE\sim \triangle BCE \Rightarrow \frac{AF}{AE}=\frac{BC}{BE}</cmath> | ||
− | By the Pythagorean Theorem, we have <math>AB=\sqrt{3^2+6^2}=3\sqrt{5}</math>, <math>AF=\sqrt{2.5^2+2.5^2}=2.5\sqrt{2}</math>, and <math>BC=\sqrt{2^2+2^2} | + | By the Pythagorean Theorem, we have <math>AB=\sqrt{3^2+6^2}=3\sqrt{5}</math>, <math>AF=\sqrt{2.5^2+2.5^2}=2.5\sqrt{2}</math>, and <math>BC=\sqrt{2^2+2^2}=2\sqrt{2}</math>. Let <math>AE=x</math>, so <math>BE=3\sqrt{5}-x</math>, then <cmath>\frac{2.5\sqrt{2}}{x}=\frac{2\sqrt{2}}{3\sqrt{5}-x}</cmath> <cmath>x=\frac{5\sqrt{5}}{3}</cmath> |
This is answer choice <math>\boxed{\text{B}}</math> | This is answer choice <math>\boxed{\text{B}}</math> | ||
− | + | ||
− | + | ==Solution 3== | |
+ | |||
+ | <asy> | ||
+ | path seg1, seg2; | ||
+ | seg1=(6,0)--(0,3); | ||
+ | seg2=(2,0)--(4,2); | ||
+ | dot((0,0)); | ||
+ | dot((1,0)); | ||
+ | fill(circle((2,0),0.1),black); | ||
+ | dot((3,0)); | ||
+ | dot((4,0)); | ||
+ | dot((5,0)); | ||
+ | fill(circle((6,0),0.1),black); | ||
+ | dot((0,1)); | ||
+ | dot((1,1)); | ||
+ | dot((2,1)); | ||
+ | dot((3,1)); | ||
+ | dot((4,1)); | ||
+ | dot((5,1)); | ||
+ | dot((6,1)); | ||
+ | dot((0,2)); | ||
+ | dot((1,2)); | ||
+ | dot((2,2)); | ||
+ | dot((3,2)); | ||
+ | fill(circle((4,2),0.1),black); | ||
+ | dot((5,2)); | ||
+ | dot((6,2)); | ||
+ | fill(circle((0,3),0.1),black); | ||
+ | dot((1,3)); | ||
+ | dot((2,3)); | ||
+ | dot((3,3)); | ||
+ | dot((4,3)); | ||
+ | dot((5,3)); | ||
+ | dot((6,3)); | ||
+ | draw(seg1); | ||
+ | draw(seg2); | ||
+ | pair [] x=intersectionpoints(seg1,seg2); | ||
+ | fill(circle(x[0],0.1),black); | ||
+ | label("$A$",(0,3),NW); | ||
+ | label("$B$",(6,0),SE); | ||
+ | label("$C$",(4,2),NE); | ||
+ | label("$D$",(2,0),S); | ||
+ | label("$E$",x[0],N); | ||
+ | label("$F$",(2,2),NE); | ||
+ | draw((2,2)--(4,2)); | ||
+ | draw((6,0)--(2,0)); | ||
+ | </asy> | ||
+ | |||
+ | Drawing line <math>\overline{BD}</math> and parallel line <math>\overline{CF}</math>, we see that <math>\triangle FCE \sim \triangle BDE</math> by AA similarity. Thus <math>\frac{FE}{EB} = \frac{FC}{DB} = \frac{2}{4} = \frac{1}{2}</math>. Reciprocating, we know that <math>\frac{EB}{FE} = 2</math> so <math>\frac{EB+FE}{FE} = 2+1 \Rightarrow \frac{FB}{FE} = 3</math>. Reciprocating again, we have <math>\frac{FE}{FB} = \frac{1}{3} \Rightarrow FE = \frac{1}{3}FB</math>. We know that <math>FD = 2</math>, so by the Pythagorean Theorem, <math>FB = \sqrt{2^{2} + 4^{2}} = 2\sqrt{5}</math>. Thus <math>FE = \frac{1}{3}FB = \frac{2\sqrt{5}}{3}</math>. Applying the Pythagorean Theorem again, we have <math>AF = \sqrt{1^{2}+2^{2}} = \sqrt{5}</math>. We finally have <math>AE = AF + FE = \sqrt{5} + \frac{2\sqrt{5}}{3} = \frac{5\sqrt{5}}{3} \Rightarrow \boxed{\text{B}}</math> | ||
+ | |||
+ | == Solution 4 == | ||
+ | |||
+ | <asy> | ||
+ | // Coordinates | ||
+ | pair A = (0,3), B = (6,0), C = (4,2), D = (2,0); | ||
+ | path seg1 = B--A; | ||
+ | path seg2 = D--C; | ||
+ | pair[] intersectionPoints = intersectionpoints(seg1, seg2); | ||
+ | pair E = intersectionPoints[0]; | ||
+ | |||
+ | for (int i = 0; i <= 6; i = i + 1) { | ||
+ | for (int j = 0; j <= 3; j = j + 1) { | ||
+ | dot((i,j)); | ||
+ | } | ||
+ | } | ||
+ | |||
+ | // Draw | ||
+ | draw(seg1); | ||
+ | draw(seg2); | ||
+ | |||
+ | dot(E); | ||
+ | |||
+ | // Label | ||
+ | label("$A$", A, NW); | ||
+ | label("$B$", B, SE); | ||
+ | label("$C$", C, dir(0)); | ||
+ | label("$D$", D, S); | ||
+ | label("$E$", E, N); | ||
+ | |||
+ | // Add extras | ||
+ | draw(C--(5,3), dashed); | ||
+ | draw(D--B, dashed); | ||
+ | draw((5,3)--A, dashed); | ||
+ | </asy> | ||
+ | |||
+ | Extend line <math>\overline{DC}</math> as above. This creates two similar triangles whose side lengths have the ratio <math>5:4</math>. Therefore <math>AE=\frac{5}{9}AB</math>. Using Pythagorean theorem to find <math>AB</math> gives us: | ||
+ | |||
+ | <cmath>AE=\frac{5}{9}AB=\frac{5}{9}\sqrt{3^2+6^2}=\frac{5}{9}\sqrt{45}= | ||
+ | \boxed{\textbf{(B) }\frac{5\sqrt{5}}{3}}</cmath> | ||
+ | |||
+ | ~ proloto | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2000|num-b=15|num-a=17}} | {{AMC10 box|year=2000|num-b=15|num-a=17}} | ||
+ | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] |
Latest revision as of 23:22, 31 July 2023
Problem
The diagram shows lattice points, each one unit from its nearest neighbors. Segment
meets segment
at
. Find the length of segment
.
Video Solution
Solution 1
Let be the line containing
and
and let
be the line containing
and
. If we set the bottom left point at
, then
,
,
, and
.
The line is given by the equation
. The
-intercept is
, so
. We are given two points on
, hence we can compute the slope,
to be
, so
is the line
Similarly, is given by
. The slope in this case is
, so
. Plugging in the point
gives us
, so
is the line
.
At , the intersection point, both of the equations must be true, so
We have the coordinates of and
, so we can use the distance formula here:
which is answer choice
Solution 2
Draw the perpendiculars from and
to
, respectively. As it turns out,
. Let
be the point on
for which
.
, and
, so by AA similarity,
By the Pythagorean Theorem, we have ,
, and
. Let
, so
, then
This is answer choice
Solution 3
Drawing line and parallel line
, we see that
by AA similarity. Thus
. Reciprocating, we know that
so
. Reciprocating again, we have
. We know that
, so by the Pythagorean Theorem,
. Thus
. Applying the Pythagorean Theorem again, we have
. We finally have
Solution 4
Extend line as above. This creates two similar triangles whose side lengths have the ratio
. Therefore
. Using Pythagorean theorem to find
gives us:
~ proloto
See Also
2000 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.