Difference between revisions of "2019 AIME II Problems/Problem 1"
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==Solution== | ==Solution== | ||
+ | <asy> | ||
+ | unitsize(10); | ||
+ | pair A = (0,0); | ||
+ | pair B = (9,0); | ||
+ | pair C = (15,8); | ||
+ | pair D = (-6,8); | ||
+ | pair E = (-6,0); | ||
+ | draw(A--B--C--cycle); | ||
+ | draw(B--D--A); | ||
+ | label("$A$",A,dir(-120)); | ||
+ | label("$B$",B,dir(-60)); | ||
+ | label("$C$",C,dir(60)); | ||
+ | label("$D$",D,dir(120)); | ||
+ | label("$E$",E,dir(-135)); | ||
+ | label("$9$",(A+B)/2,dir(-90)); | ||
+ | label("$10$",(D+A)/2,dir(-150)); | ||
+ | label("$10$",(C+B)/2,dir(-30)); | ||
+ | label("$17$",(D+B)/2,dir(60)); | ||
+ | label("$17$",(A+C)/2,dir(120)); | ||
+ | |||
+ | draw(D--E--A,dotted); | ||
+ | label("$8$",(D+E)/2,dir(180)); | ||
+ | label("$6$",(A+E)/2,dir(-90)); | ||
+ | </asy> | ||
+ | - Diagram by Brendanb4321 | ||
+ | |||
+ | |||
+ | Extend <math>AB</math> to form a right triangle with legs <math>6</math> and <math>8</math> such that <math>AD</math> is the hypotenuse and connect the points <math>CD</math> so | ||
+ | that you have a rectangle. (We know that <math>\triangle ADE</math> is a <math>6-8-10</math>, since <math>\triangle DEB</math> is an <math>8-15-17</math>.) The base <math>CD</math> of the rectangle will be <math>9+6+6=21</math>. Now, let <math>O</math> be the intersection of <math>BD</math> and <math>AC</math>. This means that <math>\triangle ABO</math> and <math>\triangle DCO</math> are with ratio <math>\frac{21}{9}=\frac73</math>. Set up a proportion, knowing that the two heights add up to 8. We will let <math>y</math> be the height from <math>O</math> to <math>DC</math>, and <math>x</math> be the height of <math>\triangle ABO</math>. | ||
+ | <cmath>\frac{7}{3}=\frac{y}{x}</cmath> | ||
+ | <cmath>\frac{7}{3}=\frac{8-x}{x}</cmath> | ||
+ | <cmath>7x=24-3x</cmath> | ||
+ | <cmath>10x=24</cmath> | ||
+ | <cmath>x=\frac{12}{5}</cmath> | ||
+ | |||
+ | This means that the area is <math>A=\tfrac{1}{2}(9)(\tfrac{12}{5})=\tfrac{54}{5}</math>. This gets us <math>54+5=\boxed{059}.</math> | ||
+ | |||
+ | -Solution by the Math Wizard, Number Magician of the Second Order, Head of the Council of the Geometers | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Using the diagram in Solution 1, let <math>E</math> be the intersection of <math>BD</math> and <math>AC</math>. We can see that angle <math>C</math> is in both | ||
+ | <math>\triangle BCE</math> and <math>\triangle ABC</math>. Since <math>\triangle BCE</math> and <math>\triangle ADE</math> are congruent by AAS, we can then state <math>AE=BE</math> and <math>DE=CE</math>. It follows that <math>BE=AE</math> and <math>CE=17-BE</math>. We can now state that the area of <math>\triangle ABE</math> is the area of <math>\triangle ABC-</math> the area of <math>\triangle BCE</math>. Using Heron's formula, we compute the area of <math>\triangle ABC=36</math>. Using the Law of Cosines on angle <math>C</math>, we obtain | ||
+ | |||
+ | <cmath>9^2=17^2+10^2-2(17)(10)cosC</cmath> | ||
+ | <cmath>-308=-340cosC</cmath> | ||
+ | <cmath>cosC=\frac{308}{340}</cmath> | ||
+ | (For convenience, we're not going to simplify.) | ||
+ | |||
+ | Applying the Law of Cosines on <math>\triangle BCE</math> yields | ||
+ | <cmath>BE^2=10^2+(17-BE)^2-2(10)(17-BE)cosC</cmath> | ||
+ | <cmath>BE^2=389-34BE+BE^2-20(17-BE)(\frac{308}{340})</cmath> | ||
+ | <cmath>0=389-34BE-(340-20BE)(\frac{308}{340})</cmath> | ||
+ | <cmath>0=389-34BE+\frac{308BE}{17}</cmath> | ||
+ | <cmath>0=81-\frac{270BE}{17}</cmath> | ||
+ | <cmath>81=\frac{270BE}{17}</cmath> | ||
+ | <cmath>BE=\frac{51}{10}</cmath> | ||
+ | This means <math>CE=17-BE=17-\frac{51}{10}=\frac{119}{10}</math>. Next, apply Heron's formula to get the area of <math>\triangle BCE</math>, which equals <math>\frac{126}{5}</math> after simplifying. Subtracting the area of <math>\triangle BCE</math> from the area of <math>\triangle ABC</math> yields the area of <math>\triangle ABE</math>, which is <math>\frac{54}{5}</math>, giving us our answer, which is <math>54+5=\boxed{059}.</math> | ||
+ | -Solution by flobszemathguy | ||
+ | |||
+ | ==Solution 3 (Very quick)== | ||
<asy> | <asy> | ||
unitsize(10); | unitsize(10); | ||
Line 24: | Line 85: | ||
label("$8$",(D+(-6,0))/2,dir(180)); | label("$8$",(D+(-6,0))/2,dir(180)); | ||
label("$6$",(A+(-6,0))/2,dir(-90)); | label("$6$",(A+(-6,0))/2,dir(-90)); | ||
+ | |||
+ | draw((4.5,0)--(4.5,2.4),dotted); | ||
+ | label("$h$", (4.5,1.2), dir(180)); | ||
+ | label("$4.5$", (6,0), dir(90)); | ||
+ | |||
</asy> | </asy> | ||
− | - Diagram by Brendanb4321 | + | - Diagram by Brendanb4321 extended by Duoquinquagintillion |
+ | |||
+ | Begin with the first step of solution 1, seeing <math>AD</math> is the hypotenuse of a <math>6-8-10</math> triangle and calling the intersection of <math>DB</math> and <math>AC</math> point <math>E</math>. Next, notice <math>DB</math> is the hypotenuse of an <math>8-15-17</math> triangle. Drop an altitude from <math>E</math> with length <math>h</math>, so the other leg of the new triangle formed has length <math>4.5</math>. Notice we have formed similar triangles, and we can solve for <math>h</math>. | ||
+ | |||
+ | <cmath>\frac{h}{4.5} = \frac{8}{15}</cmath> | ||
+ | <cmath>h = \frac{36}{15} = \frac{12}{5}</cmath> | ||
+ | |||
+ | So <math>\triangle ABE</math> has area <cmath>\frac{ \frac{12}{5} \cdot 9}{2} = \frac{54}{5}</cmath> | ||
+ | And <math>54+5=\boxed{059}.</math> | ||
+ | - Solution by Duoquinquagintillion | ||
+ | |||
+ | == Solution 4 == | ||
+ | Let <math>a = \angle{CAB}</math>. By Law of Cosines, | ||
+ | <cmath>\cos a = \frac{17^2+9^2-10^2}{2*9*17} = \frac{15}{17}</cmath> | ||
+ | <cmath>\sin a = \sqrt{1-\cos^2 a} = \frac{8}{17}</cmath> | ||
+ | <cmath>\tan a = \frac{8}{15}</cmath> | ||
+ | <cmath>A = \frac{1}{2}* 9*\frac{9}{2}\tan a = \frac{54}{5}</cmath> | ||
+ | And <math>54+5=\boxed{059}.</math> | ||
+ | |||
+ | - by Mathdummy | ||
+ | |||
+ | == Solution 5 == | ||
+ | Because <math>AD = BC</math> and <math>\angle BAD = \angle ABC</math>, quadrilateral <math>ABCD</math> is cyclic. So, Ptolemy's theorem tells us that | ||
+ | <cmath>AB \cdot CD + BC \cdot AD = AC \cdot BD \implies 9 \cdot CD + 10^2 = 17^2 \implies CD = 21.</cmath> | ||
+ | |||
+ | From here, there are many ways to finish which have been listed above. If we let <math>AB \cap CD = P</math>, then | ||
+ | <cmath>\triangle APB \sim \triangle CPD \implies \frac{AP}{AB} = \frac{CP}{CD} \implies \frac{AP}{9} = \frac{17-AP}{21} \implies AP = 5.1.</cmath> | ||
+ | |||
+ | Using Heron's formula on <math>\triangle ABP</math>, we see that | ||
+ | <cmath>[ABC] = \sqrt{9.6(9.6-5.1)(9.6-5.1)(9.6-9)} = 10.8 = \frac{54}{5}.</cmath> | ||
+ | |||
+ | Thus, our answer is <math>059</math>. ~a.y.711 | ||
+ | |||
+ | ==See Also== | ||
+ | {{AIME box|year=2019|n=II|before=First Problem|num-a=2}} | ||
+ | [[Category: Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 18:08, 26 December 2020
Contents
Problem
Two different points, and , lie on the same side of line so that and are congruent with , , and . The intersection of these two triangular regions has area , where and are relatively prime positive integers. Find .
Solution
- Diagram by Brendanb4321
Extend to form a right triangle with legs and such that is the hypotenuse and connect the points so
that you have a rectangle. (We know that is a , since is an .) The base of the rectangle will be . Now, let be the intersection of and . This means that and are with ratio . Set up a proportion, knowing that the two heights add up to 8. We will let be the height from to , and be the height of .
This means that the area is . This gets us
-Solution by the Math Wizard, Number Magician of the Second Order, Head of the Council of the Geometers
Solution 2
Using the diagram in Solution 1, let be the intersection of and . We can see that angle is in both and . Since and are congruent by AAS, we can then state and . It follows that and . We can now state that the area of is the area of the area of . Using Heron's formula, we compute the area of . Using the Law of Cosines on angle , we obtain
(For convenience, we're not going to simplify.)
Applying the Law of Cosines on yields This means . Next, apply Heron's formula to get the area of , which equals after simplifying. Subtracting the area of from the area of yields the area of , which is , giving us our answer, which is -Solution by flobszemathguy
Solution 3 (Very quick)
- Diagram by Brendanb4321 extended by Duoquinquagintillion
Begin with the first step of solution 1, seeing is the hypotenuse of a triangle and calling the intersection of and point . Next, notice is the hypotenuse of an triangle. Drop an altitude from with length , so the other leg of the new triangle formed has length . Notice we have formed similar triangles, and we can solve for .
So has area And - Solution by Duoquinquagintillion
Solution 4
Let . By Law of Cosines, And
- by Mathdummy
Solution 5
Because and , quadrilateral is cyclic. So, Ptolemy's theorem tells us that
From here, there are many ways to finish which have been listed above. If we let , then
Using Heron's formula on , we see that
Thus, our answer is . ~a.y.711
See Also
2019 AIME II (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
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.