Difference between revisions of "2021 AIME I Problems/Problem 2"
MRENTHUSIASM (talk | contribs) |
(→Video Solution) |
||
(73 intermediate revisions by 13 users not shown) | |||
Line 3: | Line 3: | ||
<asy> | <asy> | ||
pair A, B, C, D, E, F; | pair A, B, C, D, E, F; | ||
− | A = (0,3); | + | A=(0,3); |
B=(0,0); | B=(0,0); | ||
C=(11,0); | C=(11,0); | ||
Line 23: | Line 23: | ||
==Solution 1 (Similar Triangles)== | ==Solution 1 (Similar Triangles)== | ||
Let <math>G</math> be the intersection of <math>AD</math> and <math>FC</math>. | Let <math>G</math> be the intersection of <math>AD</math> and <math>FC</math>. | ||
− | From vertical angles, we know that <math>\angle FGA= \angle DGC</math>. Also, given that <math>ABCD</math> and <math>AFCE</math> are rectangles, we know that <math>\angle AFG= \angle CDG=90 ^{\circ}</math>. | + | From vertical angles, we know that <math>\angle FGA= \angle DGC</math>. Also, because we are given that <math>ABCD</math> and <math>AFCE</math> are rectangles, we know that <math>\angle AFG= \angle CDG=90 ^{\circ}</math>. |
− | Therefore, by AA similarity, we know that | + | Therefore, by AA similarity, we know that <math>\triangle AFG\sim\triangle CDG</math>. |
Let <math>AG=x</math>. Then, we have <math>DG=11-x</math>. By similar triangles, we know that <math>FG=\frac{7}{3}(11-x)</math> and <math>CG=\frac{3}{7}x</math>. We have <math>\frac{7}{3}(11-x)+\frac{3}{7}x=FC=9</math>. | Let <math>AG=x</math>. Then, we have <math>DG=11-x</math>. By similar triangles, we know that <math>FG=\frac{7}{3}(11-x)</math> and <math>CG=\frac{3}{7}x</math>. We have <math>\frac{7}{3}(11-x)+\frac{3}{7}x=FC=9</math>. | ||
Line 30: | Line 30: | ||
Solving for <math>x</math>, we have <math>x=\frac{35}{4}</math>. | Solving for <math>x</math>, we have <math>x=\frac{35}{4}</math>. | ||
The area of the shaded region is just <math>3\cdot \frac{35}{4}=\frac{105}{4}</math>. | The area of the shaded region is just <math>3\cdot \frac{35}{4}=\frac{105}{4}</math>. | ||
+ | |||
Thus, the answer is <math>105+4=\framebox{109}</math>. | Thus, the answer is <math>105+4=\framebox{109}</math>. | ||
+ | |||
~yuanyuanC | ~yuanyuanC | ||
− | ==Solution 2 (Coordinate Geometry)== | + | == Solution 2 (Similar Triangles) == |
− | Suppose <math>A=(0,0).</math> | + | |
+ | Again, let the intersection of <math>AE</math> and <math>BC</math> be <math>G</math>. By AA similarity, <math>\triangle AFG \sim \triangle CDG</math> with a <math>\frac{7}{3}</math> ratio. Define <math>x</math> as <math>\frac{[CDG]}{9}</math>. Because of similar triangles, <math>[AFG] = 49x</math>. Using <math>ABCD</math>, the area of the parallelogram is <math>33-18x</math>. Using <math>AECF</math>, the area of the parallelogram is <math>63-98x</math>. These equations are equal, so we can solve for <math>x</math> and obtain <math>x = \frac{3}{8}</math>. Thus, <math>18x = \frac{27}{4}</math>, so the area of the parallelogram is <math>33 - \frac{27}{4} = \frac{105}{4}</math>. | ||
+ | |||
+ | Finally, the answer is <math>105+4=\boxed{109}</math>. | ||
+ | |||
+ | ~mathboy100 | ||
+ | |||
+ | ==Solution 3 (Pythagorean Theorem)== | ||
+ | |||
+ | Let the intersection of <math>AE</math> and <math>BC</math> be <math>G</math>, and let <math>BG=x</math>, so <math>CG=11-x</math>. | ||
+ | |||
+ | By the Pythagorean theorem, <math>{AG}^2={AB}^2+{BG}^2</math>, so <math>AG=\sqrt{x^2+9}</math>, and thus <math>EG=9-\sqrt{x^2+9}</math>. | ||
+ | |||
+ | By the Pythagorean theorem again, <math>{CG}^2={EG}^2+{CE}^2</math>: <cmath>11-x=\sqrt{7^2+(9-\sqrt{x^2+9})^2}.</cmath> | ||
+ | |||
+ | Solving, we get <math>x=\frac{9}{4}</math>, so the area of the parallelogram is <math>3\cdot\left(11-\frac{9}{4}\right)=\frac{105}{4}</math>, and <math>105+4=\framebox{109}</math>. | ||
+ | |||
+ | ~JulianaL25 | ||
+ | |||
+ | == Solution 4 (Pythagorean Theorem)== | ||
+ | |||
+ | Let <math>P = AD \cap FC</math>, and <math>K = AE \cap BC</math>. Also let <math>AP = x</math>. | ||
+ | |||
+ | <math>CK</math> also has to be <math>x</math> by parallelogram properties. Then <math>PD</math> and <math>BK</math> must be <math>11-x</math> because the sum of the segments has to be <math>11</math>. | ||
+ | |||
+ | We can easily solve for <math>PC</math> by the Pythagorean Theorem: | ||
+ | <cmath>\begin{align*} | ||
+ | DC^2 + PD^2 &= PC^2\\ | ||
+ | 9 + (11-x)^2 &= PC^2 | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | It follows shortly that <math>PC = \sqrt{x^2-22x+30}</math>. | ||
+ | |||
+ | Also, <math>FC = 9</math>, and <math>FP + PC = 9</math>. We can then say that <math>PC = \sqrt{x^2-22x+30}</math>, so <math>FP = 9 - \sqrt{x^2-22x+30}</math>. | ||
+ | |||
+ | Now we can apply the Pythagorean Theorem to <math>\triangle AFP</math>. | ||
+ | <cmath>\begin{align*} | ||
+ | AF^2 + FP^2 = AP^2\\ | ||
+ | 49 + \left(9 - \sqrt{x^2-22x+30}\right)^2 = x^2 | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | This simplifies (not-as-shortly) to <math>x = \dfrac{35}{4}</math>. Now we have to solve for the area of <math>APCK</math>. We know that the height is <math>3</math> because the height of the parallelogram is the same as the height of the smaller rectangle. | ||
+ | |||
+ | From the area of a parallelogram (we know that the base is <math>\dfrac{35}{4}</math> and the height is <math>3</math>), it is clear that the area is <math>\dfrac{105}{4}</math>, giving an answer of <math>\boxed{109}</math>. | ||
+ | |||
+ | ~ishanvannadil2008 (Solution Sketch) | ||
+ | |||
+ | ~Tuatara (Rephrasing and <math>\LaTeX</math>) | ||
+ | |||
+ | ==Solution 5 (Coordinate Geometry)== | ||
+ | Suppose <math>B=(0,0).</math> It follows that <math>A=(0,3),C=(11,0),</math> and <math>D=(11,3).</math> | ||
+ | |||
+ | Since <math>AECF</math> is a rectangle, we have <math>AE=FC=9</math> and <math>EC=AF=7.</math> The equation of the circle with center <math>A</math> and radius <math>\overline{AE}</math> is <math>x^2+(y-3)^2=81,</math> and the equation of the circle with center <math>C</math> and radius <math>\overline{CE}</math> is <math>(x-11)^2+y^2=49.</math> | ||
+ | |||
+ | We now have a system of two equations with two variables. Expanding and rearranging respectively give | ||
+ | <cmath>\begin{align*} | ||
+ | x^2+y^2-6y&=72, &(1) \\ | ||
+ | x^2+y^2-22x&=-72. &(2) | ||
+ | \end{align*}</cmath> | ||
+ | Subtracting <math>(2)</math> from <math>(1),</math> we obtain <math>22x-6y=144.</math> Simplifying and rearranging produce <cmath>x=\frac{3y+72}{11}. \hspace{34.5mm} (*)</cmath> | ||
+ | Substituting <math>(*)</math> into <math>(1)</math> gives <cmath>\left(\frac{3y+72}{11}\right)^2+y^2-6y=72,</cmath> which is a quadratic of <math>y.</math> We clear fractions by multiplying both sides by <math>11^2=121,</math> then solve by factoring: | ||
+ | <cmath>\begin{align*} | ||
+ | \left(3y+72\right)^2+121y^2-726y&=8712 \\ | ||
+ | \left(9y^2+432y+5184\right)+121y^2-726y&=8712 \\ | ||
+ | 130y^2-294y-3528&=0 \\ | ||
+ | 2(5y+21)(13y-84)&=0 \\ | ||
+ | y&=-\frac{21}{5},\frac{84}{13}. | ||
+ | \end{align*}</cmath> | ||
+ | Since <math>E</math> is in Quadrant IV, we have <math>E=\left(\frac{3\left(-\frac{21}{5}\right)+72}{11},-\frac{21}{5}\right)=\left(\frac{27}{5},-\frac{21}{5}\right).</math> It follows that the equation of <math>\overleftrightarrow{AE}</math> is <math>y=-\frac{4}{3}x+3.</math> | ||
+ | |||
+ | Let <math>G</math> be the intersection of <math>\overline{AD}</math> and <math>\overline{FC},</math> and <math>H</math> be the intersection of <math>\overline{AE}</math> and <math>\overline{BC}.</math> Since <math>H</math> is the <math>x</math>-intercept of <math>\overleftrightarrow{AE},</math> we get <math>H=\left(\frac94,0\right).</math> | ||
− | < | + | By symmetry, quadrilateral <math>AGCH</math> is a parallelogram. Its area is <math>HC\cdot AB=\left(11-\frac94\right)\cdot3=\frac{105}{4},</math> from which the requested sum is <math>105+4=\boxed{109}.</math> |
~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | ==See | + | ==Solution 6 (Trigonometry)== |
+ | |||
+ | Let the intersection of <math>AE</math> and <math>BC</math> be <math>G</math>. It is useful to find <math>\tan(\angle DAE)</math>, because <math>\tan(\angle DAE)=\frac{3}{BG}</math> and <math>\frac{3}{\tan(\angle DAE)}=BG</math>. From there, subtracting the areas of the two triangles from the larger rectangle, we get Area = <math>33-3BG=33-\frac{9}{\tan(\angle DAE)}</math>. | ||
+ | |||
+ | let <math>\angle CAD = \alpha</math>. Let <math>\angle CAE = \beta</math>. Note, <math>\alpha+\beta=\angle DAE</math>. | ||
+ | |||
+ | <math>\alpha=\tan^{-1}\left(\frac{3}{11}\right)</math> | ||
+ | |||
+ | <math>\beta=\tan^{-1}\left(\frac{7}{9}\right)</math> | ||
+ | |||
+ | <math>\tan(\angle DAE) = \tan\left(\tan^{-1}\left(\frac{3}{11}\right)+\tan^{-1}\left(\frac{7}{9}\right)\right) = \frac{\frac{3}{11}+\frac{7}{9}}{1-\frac{3}{11}\cdot\frac{7}{9}} = \frac{\frac{104}{99}}{\frac{78}{99}} = \frac{4}{3}</math> | ||
+ | |||
+ | <math>\mathrm{Area}=33-\frac{9}{\frac{4}{3}} = 33-\frac{27}{4 } = \frac{105}{4}</math>. The answer is <math>105+4=\boxed{109}</math>. | ||
+ | |||
+ | ~twotothetenthis1024 | ||
+ | |||
+ | ==Video Solution by Punxsutawney Phil== | ||
+ | https://youtube.com/watch?v=H17E9n2nIyY&t=289s | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/M3DsERqhiDk?t=275 | ||
+ | |||
+ | ==Video Solution by Steven Chen (in Chinese)== | ||
+ | https://youtu.be/eaS5gRLSqgY | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=BinfKrc5bWo | ||
+ | |||
+ | ==Video Solution by Power of Logic== | ||
+ | https://youtu.be/WS6X1MQ37jg | ||
+ | |||
+ | ==See Also== | ||
{{AIME box|year=2021|n=I|num-b=1|num-a=3}} | {{AIME box|year=2021|n=I|num-b=1|num-a=3}} | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 23:59, 22 August 2022
Contents
- 1 Problem
- 2 Solution 1 (Similar Triangles)
- 3 Solution 2 (Similar Triangles)
- 4 Solution 3 (Pythagorean Theorem)
- 5 Solution 4 (Pythagorean Theorem)
- 6 Solution 5 (Coordinate Geometry)
- 7 Solution 6 (Trigonometry)
- 8 Video Solution by Punxsutawney Phil
- 9 Video Solution
- 10 Video Solution by Steven Chen (in Chinese)
- 11 Video Solution
- 12 Video Solution by Power of Logic
- 13 See Also
Problem
In the diagram below, is a rectangle with side lengths and , and is a rectangle with side lengths and as shown. The area of the shaded region common to the interiors of both rectangles is , where and are relatively prime positive integers. Find .
Solution 1 (Similar Triangles)
Let be the intersection of and . From vertical angles, we know that . Also, because we are given that and are rectangles, we know that . Therefore, by AA similarity, we know that .
Let . Then, we have . By similar triangles, we know that and . We have .
Solving for , we have . The area of the shaded region is just .
Thus, the answer is .
~yuanyuanC
Solution 2 (Similar Triangles)
Again, let the intersection of and be . By AA similarity, with a ratio. Define as . Because of similar triangles, . Using , the area of the parallelogram is . Using , the area of the parallelogram is . These equations are equal, so we can solve for and obtain . Thus, , so the area of the parallelogram is .
Finally, the answer is .
~mathboy100
Solution 3 (Pythagorean Theorem)
Let the intersection of and be , and let , so .
By the Pythagorean theorem, , so , and thus .
By the Pythagorean theorem again, :
Solving, we get , so the area of the parallelogram is , and .
~JulianaL25
Solution 4 (Pythagorean Theorem)
Let , and . Also let .
also has to be by parallelogram properties. Then and must be because the sum of the segments has to be .
We can easily solve for by the Pythagorean Theorem: It follows shortly that .
Also, , and . We can then say that , so .
Now we can apply the Pythagorean Theorem to .
This simplifies (not-as-shortly) to . Now we have to solve for the area of . We know that the height is because the height of the parallelogram is the same as the height of the smaller rectangle.
From the area of a parallelogram (we know that the base is and the height is ), it is clear that the area is , giving an answer of .
~ishanvannadil2008 (Solution Sketch)
~Tuatara (Rephrasing and )
Solution 5 (Coordinate Geometry)
Suppose It follows that and
Since is a rectangle, we have and The equation of the circle with center and radius is and the equation of the circle with center and radius is
We now have a system of two equations with two variables. Expanding and rearranging respectively give Subtracting from we obtain Simplifying and rearranging produce Substituting into gives which is a quadratic of We clear fractions by multiplying both sides by then solve by factoring: Since is in Quadrant IV, we have It follows that the equation of is
Let be the intersection of and and be the intersection of and Since is the -intercept of we get
By symmetry, quadrilateral is a parallelogram. Its area is from which the requested sum is
~MRENTHUSIASM
Solution 6 (Trigonometry)
Let the intersection of and be . It is useful to find , because and . From there, subtracting the areas of the two triangles from the larger rectangle, we get Area = .
let . Let . Note, .
. The answer is .
~twotothetenthis1024
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=H17E9n2nIyY&t=289s
Video Solution
https://youtu.be/M3DsERqhiDk?t=275
Video Solution by Steven Chen (in Chinese)
Video Solution
https://www.youtube.com/watch?v=BinfKrc5bWo
Video Solution by Power of Logic
See Also
2021 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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.