Difference between revisions of "2021 AIME I Problems/Problem 9"
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Call AD and BC <math>a</math>. Draw diagonal AC and call the foot of the perpendicular from B to AC <math>G</math>. Call the foot of the perpendicular from A to line BC F, and call the foot of the perpindicular from A to DC H. Triangles CBG and CAF are similar, and we get that <math>\frac{10}{15}</math>=<math>\frac{a}{AC}</math> Therefore, <math>AC=1.5a</math>. It then follows that triangles ABF and ADH are similar. Using similar triangles, we can then find that <math>AB=1.2a</math>. Using the Law of Cosine on ABC, We can find that the cosine of angle ABC is <math>-\frac{1}{3}</math>. Since angles ABF and ADH are equivalent and supplementary to angle ABC, we know that the cosine of angle ADH is 1/3. It then follows that <math>a=\frac{27\sqrt{2}}{2}</math>. Then it can be found that the area <math>K</math> is <math>\frac{567\sqrt{2}}{2}</math>. Multiplying this by <math>\sqrt{2}</math>, the answer is <math>\boxed{567}</math>. | Call AD and BC <math>a</math>. Draw diagonal AC and call the foot of the perpendicular from B to AC <math>G</math>. Call the foot of the perpendicular from A to line BC F, and call the foot of the perpindicular from A to DC H. Triangles CBG and CAF are similar, and we get that <math>\frac{10}{15}</math>=<math>\frac{a}{AC}</math> Therefore, <math>AC=1.5a</math>. It then follows that triangles ABF and ADH are similar. Using similar triangles, we can then find that <math>AB=1.2a</math>. Using the Law of Cosine on ABC, We can find that the cosine of angle ABC is <math>-\frac{1}{3}</math>. Since angles ABF and ADH are equivalent and supplementary to angle ABC, we know that the cosine of angle ADH is 1/3. It then follows that <math>a=\frac{27\sqrt{2}}{2}</math>. Then it can be found that the area <math>K</math> is <math>\frac{567\sqrt{2}}{2}</math>. Multiplying this by <math>\sqrt{2}</math>, the answer is <math>\boxed{567}</math>. | ||
-happykeeper | -happykeeper | ||
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+ | ==Solution 3(Similarity) | ||
+ | Note that all isosceles trapezoids are cyclic quadrilaterals; thus, <math>A</math> is on the circumcircle of <math>\triangle BCD</math> and we have that <math>PRQ</math> is the Simpson Line from <math>A</math>. As <math>\angle QAB = 90^\circ</math>, we have that <math>\angle QAR = 90^\circ - \angle RAB =\angle ABR = \angle APR = \angle APQ</math>, with the last equality coming from cyclic quadrilateral <math>APBR</math>. Thus, <math>\triangle QAR \sim \triangle QPA</math> and we have that <math>\frac{AQ}{AR} = \frac{PQ}{PA}</math> or that <math>\frac{18}{10} = \frac{QP}{15}</math>, which we can see gives us that <math>QP = 27</math>. Further ratios using the same similar triangles gives that <math>QR = \frac{25}{3}</math> and <math>RP = \frac{56}{3}</math>. | ||
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+ | We also see that quadrilaterals <math>APBR</math> and <math>ARDQ</math> are both cyclic, with diameters of the circumcircles being <math>AB</math> and <math>AQ</math> respectively. The intersection of the circumcircles are the points <math>A</math> and <math>R</math>, and we know <math>DRB</math> and <math>QRP</math> 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 A taking <math>\triangle APQ</math> to <math>\triangle APD</math>. Because we know a lot about <math>\triangle APQ</math> but very little about <math>\triangle APD</math> and we would like to know more, we wish to find the ratio of similitude between the two triangles. | ||
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+ | To do this, we use the one number we have for <math>\triangle APD</math>: we know that the altitude from <math>A</math> to <math>BD</math> has length 10. As the two triangles are similar, if we can find the height from <math>A</math> to <math>PQ</math>, we can take the ratio of the two heights as the ratio of similitude. To do this, we once again note that <math>QP = 27</math>. Using this, we can drop the altitude from <math>A</math> to <math>QP</math> and let it intersect <math>QP</math> at <math>H</math>. Then, let <math>QH = x</math> and thus <math>HP=27-x</math>. We then have by the Pythagorean Theorem on <math>\triangle AQH</math> and <math>\triangle APH</math>: <cmath>15^2 - x^2 = 18^2 - (27-x)^2</cmath> <cmath>225 - x^2 = 324 - (x^2-54x+729)</cmath> <cmath>54x = 630</cmath> <cmath>x=\frac{35}{3}</cmath> | ||
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+ | Then, <math>RH = QH - QR = \frac{35}{3} - \frac{25}{3} = \frac{10}{3}</math>. This gives us then from right triangle <math>\triangle ARH</math> that <math>AH = \frac{20\sqrt{2}}{3}</math> and thus the ratio of <math>\triangle APQ</math> to <math>\triangle ABD</math> is <math>\frac{3\sqrt{2}}{4}</math>. From this, we see then that <cmath>AB = AP * \frac{3\sqrt{2}}{4} = 15 * \frac{3\sqrt{2}}{4} = \frac{45\sqrt{2}}{4}</cmath> and <cmath>AD = AQ * \frac{3\sqrt{2}}{4} = 18 * \frac{3\sqrt{2}}{4} = \frac{27\sqrt{2}}{2}</cmath>. The Pythagorean Theorem on <math>\triangle AQD</math> then gives that <cmath>QD = \sqrt{AD^2 - AQ^2} = \sqrt{(\frac{27\sqrt{2}}{2})^2 - 18^2} = \sqrt{\frac{81}{2}} = \frac{9\sqrt{2}}{2}</cmath>. | ||
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+ | Then, we have the height of trapezoid <math>ABCD</math> is <math>AQ = 18</math>, the top base is <math>AB = \frac{45\sqrt{2}}{4}</math>, and the bottom base is <math>CD = \frac{45\sqrt{2}}{4} + 2*\frac{9\sqrt{2}}{2}</math>. From the equation of a trapezoid, <math>K = \frac{b_1+b_2}{2} h = \frac{63\sqrt{2}}{4} * 18 = \frac{567\sqrt{2}}{2}</math>, so the answer is <math>K\sqrt{2} = </math>\boxed{567}$. | ||
==See also== | ==See also== |
Revision as of 22:15, 12 March 2021
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
Solution 1
Construct your isosceles trapezoid. Let, for simplicity, , , and . Extend the sides and mark the intersection as . Following what the question states, drop a perpendicular from to labeling the foot as . Drop another perpendicular from to , calling the foot . Lastly, drop a perpendicular from to , labeling it . In addition, drop a perpendicular from to calling its foot .
--DIAGRAM COMING SOON--
Start out by constructing a triangle congruent to with its side of length on line . This works because all isosceles triangles are cyclic and as a result, .
Notice that by AA similarity. We are given that and by symmetry we can deduce that . As a result, . This gives us that .
The question asks us along the lines of finding the area, , of the trapezoid . We look at the area of and notice that it can be represented as . Substituting , we solve for , getting .
Now let us focus on isosceles triangle , where . Since, is an altitude from to of an isosceles triangle, must be equal to . Since and , we can solve to get that and .
We must then set up equations using the Pythagorean Theorem, writing everything in terms of , , and . Looking at right triangle we get Looking at right triangle we get Now rearranging and solving, we get two equation Those are convenient equations as which gives us After some "smart" calculation, we get that .
Notice that the question asks for , and by applying the trapezoid area formula. Fortunately, this is just , and plugging in the value of , we get that .
~Math_Genius_164
Solution 2(LOC and Trig)
Call AD and BC . Draw diagonal AC and call the foot of the perpendicular from B to AC . Call the foot of the perpendicular from A to line BC F, and call the foot of the perpindicular from A to DC H. Triangles CBG and CAF are similar, and we get that = Therefore, . It then follows that triangles ABF and ADH are similar. Using similar triangles, we can then find that . Using the Law of Cosine on ABC, We can find that the cosine of angle ABC is . Since angles ABF and ADH are equivalent and supplementary to angle ABC, we know that the cosine of angle ADH is 1/3. It then follows that . Then it can be found that the area is . Multiplying this by , the answer is . -happykeeper
==Solution 3(Similarity) Note that all isosceles trapezoids are cyclic quadrilaterals; thus, is on the circumcircle of and we have that is the Simpson 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 A 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 10. 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 \boxed{567}$.
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.