Difference between revisions of "2003 AMC 10A Problems/Problem 22"
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We extend <math>BC</math> such that it intersects <math>GF</math> at <math>X</math>. Since <math>ABCD</math> is a rectangle, it follows that <math>CD=8</math>, therefore, <math>XF=8</math>. Let <math>GX=y</math>. From the similarity of triangles <math>GCH</math> and <math>GEA</math>, we have the ratio <math>3:5</math> (as <math>CH=9-6=3</math>, and <math>EA=9-4=5</math>). <math>GX</math> and <math>GF</math> are the altitudes of <math>GCH</math> and <math>GEA</math>, respectively. Thus, <math>y:y+8 = 3:5</math>, from which we have <math>y=12</math>, thus <math>GF=y+8=12+8=\boxed{\mathrm{(B)}\ 20}</math> | We extend <math>BC</math> such that it intersects <math>GF</math> at <math>X</math>. Since <math>ABCD</math> is a rectangle, it follows that <math>CD=8</math>, therefore, <math>XF=8</math>. Let <math>GX=y</math>. From the similarity of triangles <math>GCH</math> and <math>GEA</math>, we have the ratio <math>3:5</math> (as <math>CH=9-6=3</math>, and <math>EA=9-4=5</math>). <math>GX</math> and <math>GF</math> are the altitudes of <math>GCH</math> and <math>GEA</math>, respectively. Thus, <math>y:y+8 = 3:5</math>, from which we have <math>y=12</math>, thus <math>GF=y+8=12+8=\boxed{\mathrm{(B)}\ 20}</math> | ||
− | === Solution | + | === Solution 4 === |
Since <math>GF\perp AF</math> and <math>AF\perp CD, </math> we have <math>GF\parallel CD\parallel AB.</math> Thus, <math>\triangle CDE\sim GFE.</math> Suppose <math>GF=x</math> and <math>FD=y.</math> Thus, we have <math>\dfrac{x}{8}=\dfrac{y+4}{4}.</math> Additionally, now note that <math>\triangle GAF\sim AHB,</math> which is pretty obvious from insight, but can be proven by AA with extending <math>BH</math> to meet <math>GF.</math> From this new pair of similar triangles, we have <math>\dfrac{x}{8}=\dfrac{y+9}{6}.</math> Therefore, we have by combining those two equations, <cmath>\dfrac{y+9}{6}=\dfrac{y+4}{4}.</cmath> Solving, we have <math>y=6,</math> and therefore <math>x=\boxed{\mathrm{(B)}\ 20}</math> | Since <math>GF\perp AF</math> and <math>AF\perp CD, </math> we have <math>GF\parallel CD\parallel AB.</math> Thus, <math>\triangle CDE\sim GFE.</math> Suppose <math>GF=x</math> and <math>FD=y.</math> Thus, we have <math>\dfrac{x}{8}=\dfrac{y+4}{4}.</math> Additionally, now note that <math>\triangle GAF\sim AHB,</math> which is pretty obvious from insight, but can be proven by AA with extending <math>BH</math> to meet <math>GF.</math> From this new pair of similar triangles, we have <math>\dfrac{x}{8}=\dfrac{y+9}{6}.</math> Therefore, we have by combining those two equations, <cmath>\dfrac{y+9}{6}=\dfrac{y+4}{4}.</cmath> Solving, we have <math>y=6,</math> and therefore <math>x=\boxed{\mathrm{(B)}\ 20}</math> |
Revision as of 21:53, 1 January 2020
Contents
Problem
In rectangle , we have , , is on with , is on with , line intersects line at , and is on line with . Find the length of .
Solution
Solution 1
(Vertical angles are equal).
(Both are 90 degrees).
(Alt. Interior Angles are congruent).
Therefore and are similar. and are also similar.
is 9, therefore must equal 5. Similarly, must equal 3.
Because and are similar, the ratio of and , must also hold true for and . , so is of . By Pythagorean theorem, .
.
So .
.
Therefore .
Solution 2
Since is a rectangle, .
Since is a rectangle and , .
Since is a rectangle, .
So, is a transversal, and .
This is sufficient to prove that and .
Using ratios:
Since can't have 2 different lengths, both expressions for must be equal.
Solution 3
We extend such that it intersects at . Since is a rectangle, it follows that , therefore, . Let . From the similarity of triangles and , we have the ratio (as , and ). and are the altitudes of and , respectively. Thus, , from which we have , thus
Solution 4
Since and we have Thus, Suppose and Thus, we have Additionally, now note that which is pretty obvious from insight, but can be proven by AA with extending to meet From this new pair of similar triangles, we have Therefore, we have by combining those two equations, Solving, we have and therefore
See Also
2003 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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.