Difference between revisions of "2021 Fall AMC 10A Problems/Problem 17"
MRENTHUSIASM (talk | contribs) (→Solution 3 (Extend the Sides)) |
MRENTHUSIASM (talk | contribs) (Undo revision 167668 by MRENTHUSIASM (talk)) (Tag: Undo) |
||
Line 101: | Line 101: | ||
draw(A--G, dashed); | draw(A--G, dashed); | ||
draw(C--H, dashed); | draw(C--H, dashed); | ||
− | |||
</asy> | </asy> | ||
Because of hexagon proportions, <math>\frac{BA}{AG} = \frac{1}{3}</math> and <math>\frac{BC}{CH} = \frac{1}{3}</math>. Let <math>g</math> be the height of <math>G</math>. Because <math>A</math>, <math>B</math> and <math>G</math> lie on the same line, <math>\frac{12-9}{g-12} = \frac{1}{3}</math>, so <math>g-12 = 9</math> and <math>g = 21</math>. Similarly, the height of <math>H</math> is <math>13</math>. <math>E</math> is the midpoint of <math>GH</math>, so we can take the average of these heights to get our answer, <math>\boxed{\textbf{(D) } 17}</math>. | Because of hexagon proportions, <math>\frac{BA}{AG} = \frac{1}{3}</math> and <math>\frac{BC}{CH} = \frac{1}{3}</math>. Let <math>g</math> be the height of <math>G</math>. Because <math>A</math>, <math>B</math> and <math>G</math> lie on the same line, <math>\frac{12-9}{g-12} = \frac{1}{3}</math>, so <math>g-12 = 9</math> and <math>g = 21</math>. Similarly, the height of <math>H</math> is <math>13</math>. <math>E</math> is the midpoint of <math>GH</math>, so we can take the average of these heights to get our answer, <math>\boxed{\textbf{(D) } 17}</math>. |
Revision as of 23:08, 13 December 2021
Contents
Problem
An architect is building a structure that will place vertical pillars at the vertices of regular hexagon , which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of pillars at
,
, and
are
,
, and
meters, respectively. What is the height, in meters, of the pillar at
?
Diagrams
Three-Dimensional Diagram
~MRENTHUSIASM
Two-Dimensional Diagram
~ihatemath123 ~MRENTHUSIASM
Solution 1 (Height From the Center)
The pillar at has height
and the pillar at
has height
Since the solar panel is flat, the inclination from pillar
to pillar
is
Call the center of the hexagon
Since
it follows that the solar panel has height
at
Since the solar panel is flat, the heights of the solar panel at
and
are collinear. Therefore, the pillar at
has height
~Arcticturn
Solution 2 (Height From Each Vertex)
Let the height of the pillar at be
Notice that the difference between the heights of pillar
and pillar
is equal to the difference between the heights of pillar
and pillar
So, the height at
is
Now, doing the same thing for pillar
we get the height is
Therefore, we can see the difference between the heights at pillar
and pillar
is half the difference between the heights at
and
so
The answer is
~kante314
Solution 3 (Extend the Sides)
We can extend and
to
and
, respectively, such that
and
lies on
:
Because of hexagon proportions,
and
. Let
be the height of
. Because
,
and
lie on the same line,
, so
and
. Similarly, the height of
is
.
is the midpoint of
, so we can take the average of these heights to get our answer,
.
~ihatemath123
Solution 4 (Averages of Heights)
Denote by the height of any point
.
Denote by the midpoint of
and
.
Hence,
Denote by
the center of
. Because
is a regular hexagon,
is the midpoint of
and
.
Hence,
Because
is a regular hexagon,
is the midpoint of
and
.
Hence,
Solving these equations, we get
.
~Steven Chen (www.professorchenedu.com)
Solution 5 (Vectors)
In this solution, we define rise as the change of height (in meters) from the solar panel to the ground. It follows that the rise from to
is
and the rise from
to
is
Note that
so the rise from
to
is
Together, the height of the pillar at is
meters.
~MRENTHUSIASM
Solution 6 (Vectors)
WLOG, let the side length of the hexagon be .
Establish a 3D coordinate system, in which . Let the coordinates of
and
be
,
, respectively. Then, the solar panel passes through
.
The vector and
. Computing
by the matrix
gives the result
. Therefore, a normal vector of the plane of the solar panel is
, and the equation of the plane is
. Substituting
, we find that
.
Since , we substitute
into
, which gives
.
See Also
2021 Fall AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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.