Difference between revisions of "2003 AMC 10A Problems/Problem 22"
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<math> \mathrm{(A) \ } 16\qquad \mathrm{(B) \ } 20\qquad \mathrm{(C) \ } 24\qquad \mathrm{(D) \ } 28\qquad \mathrm{(E) \ } 30 </math> | <math> \mathrm{(A) \ } 16\qquad \mathrm{(B) \ } 20\qquad \mathrm{(C) \ } 24\qquad \mathrm{(D) \ } 28\qquad \mathrm{(E) \ } 30 </math> | ||
− | == | + | == Solutions == |
=== Solution 1 === | === Solution 1 === | ||
<math>\angle GHC = \angle AHB</math> (Vertical angles are equal). | <math>\angle GHC = \angle AHB</math> (Vertical angles are equal). | ||
Line 44: | Line 44: | ||
<math>\frac{25}{10}\: =\: \frac{GF}{8}</math>. | <math>\frac{25}{10}\: =\: \frac{GF}{8}</math>. | ||
− | Therefore <math>GF= \boxed{\mathrm{(B)}\ 20}</math>. | + | Therefore <math>GF= \boxed{\mathrm{(B)}\ 20}</math>. |
=== Solution 2 === | === Solution 2 === | ||
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<math>GF=2 \cdot FD+8=2\cdot6+8=\boxed{\mathrm{(B)}\ 20}</math> | <math>GF=2 \cdot FD+8=2\cdot6+8=\boxed{\mathrm{(B)}\ 20}</math> | ||
− | === Solution 3 === | + | === Solution 3 (fastest)=== |
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> | ||
+ | |||
+ | === Solution 5 === | ||
+ | |||
+ | Since there are only lines, you can resort to coordinate bashing. Let <math>FD=k</math>. Three lines, line <math>GF</math>, line <math>GE</math>, and line <math>GA</math>, intersect at <math>G</math>. Our goal is to find the y-coordinate of that intersection point. | ||
+ | |||
+ | Line <math>GF</math> is <math>x=0</math> | ||
+ | |||
+ | Line <math>GE</math> passes through <math>(k+4, 0)</math> and <math>(k, 8)</math>. Therefore the slope is <math>-2</math> and the line is | ||
+ | <math>y-0=-2(x-k-4)</math> which is <math>y=-2x+2k+8</math> | ||
+ | |||
+ | Line <math>GA</math> passes through <math>(k+9, 0)</math> and <math>(k+3, 8)</math>. Therefore the slope is <math>\frac{-4}{3}</math> and the line is | ||
+ | <math>y-0=-\frac{4}{3}(x-k-9)</math> which simplifies to <math>y=-\frac{4}{3}x+\frac{4}{3}k+12</math> | ||
+ | |||
+ | We solve the system of equations with these three lines. First we plug in <math>x=0</math> | ||
+ | |||
+ | <math>y=2k+8</math> | ||
+ | |||
+ | <math>y=\frac{4}{3}k+12</math> | ||
+ | |||
+ | Next, we solve for k. <math>k=6</math> Therefore <math>y=20</math>. The y-coordinate of this intersection point is indeed our answer. <math>\boxed{\mathrm{(B)}\ 20}</math> | ||
+ | ~superagh | ||
+ | |||
+ | === Solution 6 (simple coordinates)=== | ||
+ | Let <math>A</math> be the origin of our coordinate system. Now line <math>GA</math> has equation <math>-\frac{4}{3}x</math>. We can use point-slope form to find the equation for line <math>GE</math>. First, we know that its slope is <math>-2</math>, and we know that it passes through <math>E=(-5,0)</math>, so line <math>GE</math> has equation <math>-2(x+5)</math>. Solving for the intersection by letting <math>-\frac{4}{3}x=-2(x+5)</math>, we get x=-15. Plugging this into our equation for line <math>GA</math> gives us <math>G=(-15,20)</math>, so <math>GF= \boxed{\mathrm{(B)}\ 20}</math> | ||
+ | ~chrisdiamond10 | ||
== See Also == | == See Also == |
Latest revision as of 23:30, 4 January 2021
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 .
Solutions
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 (fastest)
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
Solution 5
Since there are only lines, you can resort to coordinate bashing. Let . Three lines, line , line , and line , intersect at . Our goal is to find the y-coordinate of that intersection point.
Line is
Line passes through and . Therefore the slope is and the line is which is
Line passes through and . Therefore the slope is and the line is which simplifies to
We solve the system of equations with these three lines. First we plug in
Next, we solve for k. Therefore . The y-coordinate of this intersection point is indeed our answer. ~superagh
Solution 6 (simple coordinates)
Let be the origin of our coordinate system. Now line has equation . We can use point-slope form to find the equation for line . First, we know that its slope is , and we know that it passes through , so line has equation . Solving for the intersection by letting , we get x=-15. Plugging this into our equation for line gives us , so ~chrisdiamond10
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 |
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