Difference between revisions of "2009 AMC 10A Problems/Problem 17"
(New page: == Problem == Rectangle <math>ABCD</math> has <math>AB=4</math> and <math>BC=3</math>. Segment <math>EF</math> is constructed through <math>B</math> so that <math>EF</math> is perpendicula...) |
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\mathrm{(E)}\ 12 | \mathrm{(E)}\ 12 | ||
</math> | </math> | ||
+ | [[Category: Introductory Geometry Problems]] | ||
− | == Solution == | + | == Solutions == |
+ | |||
+ | === Solution 1 === | ||
The situation is shown in the picture below. | The situation is shown in the picture below. | ||
Line 40: | Line 43: | ||
</asy> | </asy> | ||
− | + | From the [[Pythagorean theorem]] we have <math>BD=5</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>mBDC = mDBA</math> | ||
+ | |||
+ | <math>mADB + mBDC = mBDC + mABE</math> | ||
+ | |||
+ | <math>mADB = mABE</math> | ||
+ | |||
+ | Next, because every triangle has a degree measure of <math>180</math>, angle <math>BEA</math> and angle <math>DBA</math> are similar. | ||
+ | |||
− | + | Hence <math>BE/AB = DB/AD</math>, and therefore <math>BE = AB\cdot DB/AD = 20/3</math>. | |
Also triangle <math>CBF</math> is similar to <math>ABD</math>. Hence <math>BF/BC = DB/AB</math>, and therefore <math>BF=BC\cdot DB / AB = 15/4</math>. | Also triangle <math>CBF</math> is similar to <math>ABD</math>. Hence <math>BF/BC = DB/AB</math>, and therefore <math>BF=BC\cdot DB / AB = 15/4</math>. | ||
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We then have <math>EF = EB+BF = \frac{20}3 + \frac{15}4 = \frac{80 + 45}{12} = \boxed{\frac{125}{12}}</math>. | We then have <math>EF = EB+BF = \frac{20}3 + \frac{15}4 = \frac{80 + 45}{12} = \boxed{\frac{125}{12}}</math>. | ||
+ | === Solution 2 === | ||
+ | |||
+ | 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>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 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 === | ||
+ | There is a better solution where we find <math>EF</math> directly instead of in parts (use similarity). The strategy is similar. | ||
+ | |||
+ | ===Solution 4(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>. | ||
+ | Binderclips1 01:04, 10 June 2020 (EDT) | ||
== 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}} |
Latest revision as of 10:47, 3 August 2021
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
There is a better solution where we find directly instead of in parts (use similarity). The strategy is similar.
Solution 4(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 = . Binderclips1 01:04, 10 June 2020 (EDT)
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.