Difference between revisions of "2016 AMC 10B Problems/Problem 19"
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==Solution 3 (Similar Triangles)== | ==Solution 3 (Similar Triangles)== | ||
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pair A1=(2,0),A2=(4,4); | pair A1=(2,0),A2=(4,4); | ||
pair B1=(0,4),B2=(5,1); | pair B1=(0,4),B2=(5,1); | ||
Line 104: | Line 104: | ||
dot(H); | dot(H); | ||
dot((0,0)); dot((5,4)); | dot((0,0)); dot((5,4)); | ||
− | + | <\asy> | |
Extend <math>AG</math> to intersect <math>CD</math> at <math>H</math>. Letting <math>x=\overline{HC}</math>, we have that <cmath>\triangle{HCG}\sim\triangle{HDA}\implies \dfrac{\overline{HC}}{\overline{CG}}=\dfrac{\overline{HD}}{\overline{AD}}\implies \dfrac{x}{1}=\dfrac{x+5}{4}\implies x=\dfrac{5}{3}.</cmath> | Extend <math>AG</math> to intersect <math>CD</math> at <math>H</math>. Letting <math>x=\overline{HC}</math>, we have that <cmath>\triangle{HCG}\sim\triangle{HDA}\implies \dfrac{\overline{HC}}{\overline{CG}}=\dfrac{\overline{HD}}{\overline{AD}}\implies \dfrac{x}{1}=\dfrac{x+5}{4}\implies x=\dfrac{5}{3}.</cmath> |
Revision as of 16:44, 21 February 2016
Contents
Problem
Rectangle has
and
. Point
lies on
so that
, point
lies on
so that
. and point
lies on
so that
. Segments
and
intersect
at
and
, respectively. What is the value of
?
Solution 1 (Answer Choices)
Since the opposite sides of a rectangle are parallel and
due to vertical angles,
. Furthermore, the ratio between the side lengths of the two triangles is
. Labeling
and
, we see that
turns out to be equal to
. Since the denominator of
must now be a multiple of 7, the only possible solution in the answer choices is
.
Solution 2 (Coordinate Geometry)
First, we will define point as the origin. Then, we will find the equations of the following three lines:
,
, and
. The slopes of these lines are
,
, and
, respectively. Next, we will find the equations of
,
, and
. They are as follows:
After drawing in altitudes to
from
,
, and
, we see that
because of similar triangles, and so we only need to find the x-coordinates of
and
.
Finding the intersections of
and
, and
and
gives the x-coordinates of
and
to be
and
. This means that
. Now we can find
Solution 3 (Similar Triangles)
<asy>
pair A1=(2,0),A2=(4,4);
pair B1=(0,4),B2=(5,1);
pair C1=(5,0),C2=(0,4);
pair H = (20/3,0);
draw(A1--A2);
draw(B1--B2);
draw(C1--C2);
draw(B1--H);
draw((0,0)--H);
draw((0,0)--B1--(5,4)--C1--cycle);
dot((20/7,12/7));
dot((3.07692307692,2.15384615384));
label("",(3.07692307692,2.15384615384),N);
label("
",(20/7,12/7),W);
label("
",(0,4), NW);
label("
",(5,4), NE);
label("
",(5,0),SE);
label("
",(0,0),SW);
label("
",(2,0),S); label("
",(5,1),E);
label("
",(4,4),N);
label("
",H,E);
dot(A1); dot(A2); dot(B1); dot(B2); dot(C1); dot(C2); dot(H); dot((0,0)); dot((5,4)); <\asy>
Extend to intersect
at
. Letting
, we have that
Then, notice that and
. Thus, we see that
and
Thus, we see that
See Also
2016 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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.