Difference between revisions of "2009 AMC 10A Problems/Problem 17"
m (→[Solution 1]) |
(→Solution 3(Coordinate Bash)) |
||
(13 intermediate revisions by 8 users not shown) | |||
Line 45: | Line 45: | ||
From the [[Pythagorean theorem]] we have <math>BD=5</math>. | From the [[Pythagorean theorem]] we have <math>BD=5</math>. | ||
− | Triangle <math>EAB</math> is similar to <math> | + | Triangle <math>EAB</math> is similar to <math>BAD</math>, as they have the same angles. Segment <math>BA</math> is perpendicular to <math>DA</math>, meaning that angle <math>DAB</math> and <math>BAE</math> are right angles and congruent. Also, angle <math>DBE</math> is a right angle. Because it is a rectangle, angle <math>BDC</math> is congruent to <math>DBA</math> and angle <math>ADC</math> is also a right angle. By the transitive property: |
<math>mADB + mBDC = mDBA + mABE</math> | <math>mADB + mBDC = mDBA + mABE</math> | ||
Line 55: | Line 55: | ||
<math>mADB = mABE</math> | <math>mADB = mABE</math> | ||
− | Next, because every triangle has a degree measure of 180, angle <math> | + | Next, because every triangle has a degree measure of <math>180</math>, angle <math>BEA</math> and angle <math>DBA</math> are similar. |
Line 66: | Line 66: | ||
=== Solution 2 === | === Solution 2 === | ||
− | Since <math>BD</math> is the altitude from <math> | + | Since <math>BD</math> is the altitude from <math>D</math> to <math>EF</math>, we can use the equation <math>BD^2 = EB\cdot BF</math>. |
− | Looking at the angles, we see that triangle <math> | + | Looking at the angles, we see that triangle <math>BDE</math> is similar to <math>DCB</math>. Because of this, <math>\frac{AB}{CB} = \frac{EB}{DB}</math>. From the given information and the [[Pythagorean theorem]], <math>AB=4</math>, <math>CB=3</math>, and <math>DB=5</math>. Solving gives <math>EB=20/3</math>. |
We can use the above formula to solve for <math>BF</math>. <math>BD^2 = 20/3\cdot BF</math>. Solve to obtain <math>BF=15/4</math>. | We can use the above formula to solve for <math>BF</math>. <math>BD^2 = 20/3\cdot BF</math>. Solve to obtain <math>BF=15/4</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>. | 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(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 <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> | ||
+ | is on the line <math>y=\frac{-4x}{3}+\frac{25}{3}</math>. Since <math>E</math> and <math>F</math> are on the <math>y</math> and <math>x</math> axis, respectively, we plug in <math>0</math> for <math>x</math> | ||
+ | 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>. | ||
== See Also == | == See Also == |
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.