2021 AIME I Problems/Problem 9
Contents
- 1 Problem
- 2 Diagram
- 3 Solution 1 (Similar Triangles and Pythagorean Theorem)
- 4 Solution 2 (Similar Triangles and Pythagorean Theorem)
- 5 Solution 3 (Similar Triangles and Pythagorean Theorem)
- 6 Solution 4 (Cyclic Quadrilaterals, Similar Triangles, Pythagorean Theorem)
- 7 Solution 5 (Cyclic Quadrilaterals, Similar Triangles, Pythagorean Theorem)
- 8 Solution 6 (Similar Triangles and Trigonometry)
- 9 Solution 7 (Similar Triangles and Trigonometry)
- 10 Solution 8 (Heron's Formula)
- 11 Solution 9 (Three Heights)
- 12 Solution 10 (Area)
- 13 Video Solution
- 14 Video Solution
- 15 See Also
Problem
Let be an isosceles trapezoid with and Suppose that the distances from to the lines and are and respectively. Let be the area of Find
Diagram
~MRENTHUSIASM
Solution 1 (Similar Triangles and Pythagorean Theorem)
Let and be the perpendiculars from to and respectively. Next, let be the intersection of and
We set and as shown below. From here, we obtain by segment subtraction, and and by the Pythagorean Theorem.
Since and are both complementary to we have from which by AA. It follows that so or Since by vertical angles, we have by AA, with the ratio of similitude It follows that
Since by angle chasing, we have by AA, with the ratio of similitude It follows that
By the Pythagorean Theorem on right we have or Solving this system of equations ( and ), we get and so and Finally, the area of is from which
~MRENTHUSIASM
Remark
Instead of solving the system of equations and which can be time consuming, by noting that by AA, we could find out , which gives . We also know that by Pythagorean Theorem on . From we apply the Pythagorean Theorem to and obtain Substituting, we get from which
~Chupdogs
Solution 2 (Similar Triangles and Pythagorean Theorem)
First, draw the diagram. Then, notice that since is isosceles, , and the length of the altitude from to is also . Let the foot of this altitude be , and let the foot of the altitude from to be denoted as . Then, . So, . Now, notice that , where denotes the area of triangle . Letting , this equality becomes . Also, from , we have . Now, by the Pythagorean theorem on triangles and , we have and . Notice that , so . Squaring both sides of the equation once, moving and to the right, dividing both sides by , and squaring the equation once more, we are left with . Dividing both sides by (since we know is positive), we are left with . Solving for gives us .
Now, let the foot of the perpendicular from to be . Then let . Let the foot of the perpendicular from to be . Then, is also equal to . Notice that is a rectangle, so . Now, we have . By the Pythagorean theorem applied to , we have . We know that , so we can plug this into this equation. Solving for , we get .
Finally, to find , we use the formula for the area of a trapezoid: . The problem asks us for , which comes out to be .
~advanture
Solution 3 (Similar Triangles and Pythagorean Theorem)
Make perpendicular to ; perpendicular to ; perpendicular .
It's obvious that . Let . Then make perpendicular to , it's easy to get .
Since parallel to , , so . After drawing the altitude, it's obvious that , so . According to the property of similar triangles, . So, , or .
Now, we see the , pretty easy to find that , then we get , then express into form that we put the length of back to : . So, After calculating, we can have a final equation of . It's easy to find then . So, ~bluesoul
Solution 4 (Cyclic Quadrilaterals, Similar Triangles, Pythagorean Theorem)
Let the foot of the altitude from to be , to be , and to be .
Note that all isosceles trapezoids are cyclic quadrilaterals; thus, is on the circumcircle of and we have that is the Simson Line from . As , we have that , with the last equality coming from cyclic quadrilateral . Thus, and we have that or that , which we can see gives us that . Further ratios using the same similar triangles gives that and .
We also see that quadrilaterals and are both cyclic, with diameters of the circumcircles being and respectively. The intersection of the circumcircles are the points and , and we know and are both line segments passing through an intersection of the two circles with one endpoint on each circle. By Fact 5, we know then that there exists a spiral similarity with center taking to . Because we know a lot about but very little about and we would like to know more, we wish to find the ratio of similitude between the two triangles.
To do this, we use the one number we have for : we know that the altitude from to has length . As the two triangles are similar, if we can find the height from to , we can take the ratio of the two heights as the ratio of similitude. To do this, we once again note that . Using this, we can drop the altitude from to and let it intersect at . Then, let and thus . We then have by the Pythagorean Theorem on and : Then, . This gives us then from right triangle that and thus the ratio of to is . From this, we see then that and The Pythagorean Theorem on then gives that Then, we have the height of trapezoid is , the top base is , and the bottom base is . From the equation of a trapezoid, , so the answer is .
~lvmath
Solution 5 (Cyclic Quadrilaterals, Similar Triangles, Pythagorean Theorem)
Let and be the feet of the altitudes from to and , respectively.
Claim: We have pairs of similar right triangles: and .
Proof: Note that is cyclic. We need one more angle, and we get this from this cyclic quadrilateral: Let . We obtain from the similarities and .
By Ptolemy, , so .
We obtain , so .
Applying the Pythagorean theorem on , we get .
Thus, , and , yielding .
Solution 6 (Similar Triangles and Trigonometry)
Let . Draw diagonal and let be the foot of the perpendicular from to , be the foot of the perpendicular from to line , and be the foot of the perpendicular from to .
Note that , and we get that . Therefore, . It then follows that . Using similar triangles, we can then find that . Using the Law of Cosines on , We can find that the . Since , and each is supplementary to , we know that the . It then follows that . Then it can be found that the area is . Multiplying this by , the answer is .
~happykeeper
Solution 7 (Similar Triangles and Trigonometry)
Draw the distances in terms of , as shown in the diagram. By similar triangles, . As a result, let , then and . The triangle is which . By angle subtraction, . Therefore, and . By trapezoid area formula, the area of is equal to which .
~math2718281828459
Solution 8 (Heron's Formula)
Let the points formed by dropping altitudes from to the lines , , and be , , and , respectively.
We have and For convenience, let . By Heron's formula on , we have sides and semiperimeter , so so .
Then, and Finally, recalling that is isosceles, so .
Solution 9 (Three Heights)
Let and be the perpendiculars from to and respectively. . Denote by the base of the perpendicular from to be the base of the perpendicular from to . Denote It is clear that the area of is equal to the area of the rectangle
The problem is reduced to finding .
In triangle all altitudes are known: We apply the Law of Cosines to and get We apply the Pythagorean Law to and get Required area is
vladimir.shelomovskii@gmail.com, vvsss
Solution 10 (Area)
Let be on such that . Let be on such that .
Let be the length of . Let be the length of .
The area of can be expressed in three ways: , , and .
Now, . We can substitute in to get
.
We have After a fairly straightforward algebraic bash, we get , and . By the Pythagorean Theorem on , , and .
Thus, . Therefore, . The requested answer is .
~ adam_zheng
Video Solution
~Mathproblemsolvingskills.com
Video Solution
https://www.youtube.com/watch?v=6rLnl8z7lnM
See Also
2021 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.