Difference between revisions of "2009 AMC 10A Problems/Problem 17"
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We now know <math>EB</math> and <math>BF</math>. <math>EF = EB+BF = \frac{20}3 + \frac{15}4 = \frac{80 + 45}{12} = \boxed{\frac{125}{12}}</math>. | We now know <math>EB</math> and <math>BF</math>. <math>EF = EB+BF = \frac{20}3 + \frac{15}4 = \frac{80 + 45}{12} = \boxed{\frac{125}{12}}</math>. | ||
− | === Solution 3 | + | ===Solution 3(Coordinate Bash)=== |
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To keep things simple, we will use coordinates in only the first quadrant. The picture will look like the diagram above reflected over the <math>x</math>-axis.It is also worth noting the <math>F</math> will lie on the <math>x</math> axis and <math>E</math> on the <math>y</math>. Let <math>D</math> be the origin, <math>A(3,0)</math>, <math>C(4,0)</math>, and <math>B(4,3)</math>. We can express segment <math>DB</math> as the line <math>y=\frac{3x}{4}</math>. | To keep things simple, we will use coordinates in only the first quadrant. The picture will look like the diagram above reflected over the <math>x</math>-axis.It is also worth noting the <math>F</math> will lie on the <math>x</math> axis and <math>E</math> on the <math>y</math>. Let <math>D</math> be the origin, <math>A(3,0)</math>, <math>C(4,0)</math>, and <math>B(4,3)</math>. We can express segment <math>DB</math> as the line <math>y=\frac{3x}{4}</math>. | ||
Since <math>EF</math> is perpendicular to <math>DB</math>, and we know that <math>(4,3)</math> lies on it, we can use this information to find that segment <math>EF</math> | Since <math>EF</math> is perpendicular to <math>DB</math>, and we know that <math>(4,3)</math> lies on it, we can use this information to find that segment <math>EF</math> | ||
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and <math>y</math>, we find that point <math>E</math> is at <math>(0,\frac{25}{3})</math>, and point <math>F</math> is at <math>(\frac{25}{4},0)</math>. Applying the distance formula, | and <math>y</math>, we find that point <math>E</math> is at <math>(0,\frac{25}{3})</math>, and point <math>F</math> is at <math>(\frac{25}{4},0)</math>. Applying the distance formula, | ||
we obtain that <math>EF</math>= <math>\boxed{\frac{125}{12}}</math>. | we obtain that <math>EF</math>= <math>\boxed{\frac{125}{12}}</math>. | ||
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== See Also == | == See Also == | ||
{{AMC10 box|year=2009|ab=A|num-b=16|num-a=18}} | {{AMC10 box|year=2009|ab=A|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 21:14, 23 October 2023
Contents
Problem
Rectangle has
and
. Segment
is constructed through
so that
is perpendicular to
, and
and
lie on
and
, respectively. What is
?
Solutions
Solution 1
The situation is shown in the picture below.
From the Pythagorean theorem we have .
Triangle is similar to
, as they have the same angles. Segment
is perpendicular to
, meaning that angle
and
are right angles and congruent. Also, angle
is a right angle. Because it is a rectangle, angle
is congruent to
and angle
is also a right angle. By the transitive property:
Next, because every triangle has a degree measure of , angle
and angle
are similar.
Hence , and therefore
.
Also triangle is similar to
. Hence
, and therefore
.
We then have .
Solution 2
Since is the altitude from
to
, we can use the equation
.
Looking at the angles, we see that triangle is similar to
. Because of this,
. From the given information and the Pythagorean theorem,
,
, and
. Solving gives
.
We can use the above formula to solve for .
. Solve to obtain
.
We now know and
.
.
Solution 3(Coordinate Bash)
To keep things simple, we will use coordinates in only the first quadrant. The picture will look like the diagram above reflected over the -axis.It is also worth noting the
will lie on the
axis and
on the
. Let
be the origin,
,
, and
. We can express segment
as the line
.
Since
is perpendicular to
, and we know that
lies on it, we can use this information to find that segment
is on the line
. Since
and
are on the
and
axis, respectively, we plug in
for
and
, we find that point
is at
, and point
is at
. Applying the distance formula,
we obtain that
=
.
See Also
2009 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.