2009 AMC 10A Problems/Problem 17
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 | |
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