Difference between revisions of "2000 AMC 10 Problems/Problem 16"
Shefroggie (talk | contribs) m (→Solution 2) |
Flyhawkeye (talk | contribs) (Added solution 3) |
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Also, you could extend CD to the end of the box and create two similar triangles. Then use ratios and find that the distance is 5/9 of the diagonal AB. Thus, the answer is B. | Also, you could extend CD to the end of the box and create two similar triangles. Then use ratios and find that the distance is 5/9 of the diagonal AB. Thus, the answer is B. | ||
+ | |||
+ | ===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((6,2)--(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 by AA similarity. Thus </math>\frac{FE}{EB} = \frac{FC}{DB} = \frac{2}{4} = \frac{1}{2}<math></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 = </math>\sqrt{5} + \frac{2\sqrt{5}}{3}<math> = \frac{5\sqrt{5}}{3} \Rightarrow \boxed{\text{B}}</math> | ||
==See Also== | ==See Also== |
Revision as of 18:35, 29 October 2017
Problem
The diagram shows lattice points, each one unit from its nearest neighbors. Segment
meets segment
at
. Find the length of segment
.
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
Also, you could extend CD to the end of the box and create two similar triangles. Then use ratios and find that the distance is 5/9 of the diagonal AB. Thus, the answer is B.
Solution 3
Drawing line and parallel line
, we see that
\frac{FE}{EB} = \frac{FC}{DB} = \frac{2}{4} = \frac{1}{2}$$ (Error compiling LaTeX. Unknown error_msg). 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
\sqrt{5} + \frac{2\sqrt{5}}{3}
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.