Difference between revisions of "2021 AIME I Problems/Problem 2"
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MRENTHUSIASM (talk | contribs) (As we discourage bashing, I have moved the solutions around based on elegance. Let me know if you disagree.) |
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~yuanyuanC | ~yuanyuanC | ||
− | ==Solution 2 | + | ==Solution 2 (Pythagorean Theorem)== |
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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>. | 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>. | ||
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~JulianaL25 | ~JulianaL25 | ||
− | == Solution | + | == Solution 3 (Similar Triangles and Area) == |
Again, let the intersection of <math>AE</math> and <math>BC</math> be <math>G</math>. By AA similarity, <math>AFG \sim 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>. | Again, let the intersection of <math>AE</math> and <math>BC</math> be <math>G</math>. By AA similarity, <math>AFG \sim 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>. | ||
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~mathboy100 | ~mathboy100 | ||
− | == Solution | + | == Solution 4 == |
Let <math>P = AD \cap FC</math>, and <math>K = AE \cap BC</math>. Also let <math>AP = x</math>. | Let <math>P = AD \cap FC</math>, and <math>K = AE \cap BC</math>. Also let <math>AP = x</math>. | ||
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~Tuatara (Rephrasing and <math>\LaTeX</math>) | ~Tuatara (Rephrasing and <math>\LaTeX</math>) | ||
+ | |||
+ | ==Solution 5 (Coordinate Geometry Bash)== | ||
+ | Suppose <math>B=(0,0).</math> It follows that | ||
+ | <cmath>\begin{align*} | ||
+ | A&=(0,3), \\ | ||
+ | C&=(11,0), \\ | ||
+ | D&=(11,3). | ||
+ | \end{align*}</cmath> | ||
+ | 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 get <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(5x+21)(13x-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 obtain <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 | ||
==Solution 6 (Trigonometry Bash)== | ==Solution 6 (Trigonometry Bash)== |
Revision as of 02:54, 8 September 2021
Contents
- 1 Problem
- 2 Solution 1 (Similar Triangles)
- 3 Solution 2 (Pythagorean Theorem)
- 4 Solution 3 (Similar Triangles and Area)
- 5 Solution 4
- 6 Solution 5 (Coordinate Geometry Bash)
- 7 Solution 6 (Trigonometry Bash)
- 8 Video Solution by Punxsutawney Phil
- 9 Video Solution
- 10 Video Solution by Steven Chen (in Chinese)
- 11 Video Solution
- 12 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 triangles and are similar.
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 (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 3 (Similar Triangles and Area)
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 4
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 Bash)
Suppose It follows that 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 get 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 obtain
By symmetry, quadrilateral is a parallelogram. Its area is from which the requested sum is
~MRENTHUSIASM
Solution 6 (Trigonometry Bash)
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
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.