2003 AMC 10A Problems/Problem 22
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 . Plugging this into our equation for line gives us , so ~chrisdiamond10
Solution 7 (system of equations through angle similarity)
First, using given information, we can find the values of some line segments in the figure. We find that (through Pythagorean Theorem), , and . Let Line and let Line . We find that through some angle chasing (they both have a right angle, and they both share angle . Using this information, we can write the equation . Through simplifying this equation, we get that . Let point be the point on line so that lines and are perpendicular, and we get that and . Doing some more angle chasing, we can find that , as they both share and they both have a right angle.
With this information, we can write the equation Simplifying this equation we get the equation . Plugging in for , we get , so . Lastly, to find the value of y, which is the value of Line , our desired value, we plug in for in the equation , we get , which, finally, we get our value of , so therefore, our answer is
~Darth_Cadet
Solution 8 Line and Slope
Draw a coordinate plane with the y-axis centered on and the x-axis centered on . From there, call the line passing through and the line passing through . From there, you can find that the equations for these lines are and respectively. We need to find the length of , so we are finding the y-value of the point . Solving for this point, we get ;;. Now, plugging in the values, we find that , so our answer is .
~iamcalifornia'sresidentidiot
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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