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)); | |
− | draw( | + | </asy> |
− | label(" | + | - 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> | ||
+ | unitsize(10); | ||
+ | pair A = (0,0); | ||
+ | pair B = (9,0); | ||
+ | pair C = (15,8); | ||
+ | pair D = (-6,8); | ||
+ | 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("$9$",(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--(-6,0)--A,dotted); | ||
+ | label("$8$",(D+(-6,0))/2,dir(180)); | ||
+ | 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> | ||
+ | - 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 | ||
+ | |||
+ | == Solution 6 == | ||
+ | |||
+ | Let <math>A=(0,0), B=(9,0)</math>. Now consider <math>C</math>, and if we find the coordinates of <math>C</math>, by symmetry about <math>x=4.5</math>, we can find the coordinates of D. | ||
+ | |||
+ | So let <math>C=(a,b)</math>. So the following equations hold: | ||
+ | |||
+ | <math>\sqrt{(a-9)^2+(b)^2}=17</math>. | ||
+ | |||
+ | <math>\sqrt{a^2+b^2}=10</math>. | ||
+ | |||
+ | Solving by squaring both equations and then subtracting one from the other to eliminate <math>b^2</math>, we get <math>C=(-6,8)</math> because <math>C</math> is in the second quadrant. | ||
+ | |||
+ | Now by symmetry, <math>D=(16, 8)</math>. | ||
+ | |||
+ | So now you can proceed by finding the intersection and then calculating the area directly. We get <math>\boxed{059}</math>. | ||
+ | |||
+ | ~hastapasta | ||
+ | |||
+ | == Solution 7 == | ||
+ | |||
+ | Since the figure formed by connecting the vertices of the congruent triangles is a isoceles trapezoid, by Ptolemys, the other base of the trapezoid is <math>21.</math> Then, dropping altitudes to the base of <math>21</math> and using pythagorean theorem, we have the height is <math>8,</math> and we can use similar triangles to finish. | ||
+ | |||
+ | == Solution 8 (Very, ''very'', quick, but for observant people only) == | ||
+ | |||
+ | <asy> | ||
+ | //Made by Afly. I used some resources. | ||
+ | //Took me 10 min to get everything right. | ||
+ | import olympiad; | ||
+ | unitsize(18); | ||
+ | pair A = (0,0); | ||
+ | pair B = (0,8); | ||
+ | pair C = (6,0); | ||
+ | pair D = (15,0); | ||
+ | pair E = (21,0); | ||
+ | pair F = (21,8); | ||
+ | pair G = (21/2,0); | ||
+ | pair H = intersectionpoints(B--D,C--F)[0]; | ||
+ | pen dash1 = linetype(new real [] {9,9})+linewidth(1); | ||
+ | pen solid1 = linetype(new real [] {9,0})+linewidth(1); | ||
+ | pen dash2 = linetype(new real [] {3,3})+linewidth(1); | ||
+ | fill(C--G--H--cycle,rgb(3/4,1/4,1/4)); | ||
+ | fill(D--G--H--cycle,rgb(3/4,3/4,1/4)); | ||
+ | draw(C--A--B,dash1); | ||
+ | draw(C--B--D--C,solid1); | ||
+ | draw(F--E--D,dash1); | ||
+ | draw(F--D--C--F,solid1); | ||
+ | draw(G--H,dash2); | ||
+ | draw(brace(D+dir(270),A+dir(270)),solid1); | ||
+ | draw(brace(D,C),solid1); | ||
+ | draw(A--A+2*dir(180),dash1,EndArrow); | ||
+ | draw(E--E+2*dir(0),dash1,EndArrow); | ||
+ | pair L1 = (15/2,-7/2); | ||
+ | pair L2 = (21/2,-13/8); | ||
+ | label("15",L1); | ||
+ | label("8",A--B,W); | ||
+ | label("6",A--C,S); | ||
+ | label("10",B--C,SW); | ||
+ | label("17",B--D,NE); | ||
+ | label("9",L2); | ||
+ | label("4.5",G--D,S); | ||
+ | label("2.4",G--H,W); | ||
+ | markscalefactor = 1/16; | ||
+ | draw(rightanglemark(H,G,D)); | ||
+ | draw(rightanglemark(B,A,C)); | ||
+ | draw(rightanglemark(D,E,F)); | ||
+ | label("A",C,SW); | ||
+ | label("B",D,SE); | ||
+ | label("C",B,NW); | ||
+ | label("D",F,NE); | ||
+ | label("E",A,SW); | ||
+ | label("F",E,SE); | ||
+ | label("G",G,NW); | ||
+ | label("H",H,N); | ||
+ | </asy> | ||
+ | |||
+ | First, let's define H as the intersection of CB and DA, and define G as the midpoint of AB. Next, let E and F be the feet of the perpendicular lines from C and D to line AB respectively. We get a pleasing line of symmetry HG where A maps to B, C maps to D, and E maps to F. We notice that 8-15-17 and 8-6-10 are both pythagorean triples and we test that theory. Then, we find AB = 15 - 6 = 9 and GB = 4.5. Since △CEB~△HGB, we find that HG = 2.4. Then, the area of △HGB is 5.4 and the total overlap is 10.8 = 54/5. Noting that GCD(54,5)=1, we add them to get 59. | ||
+ | |||
+ | Note: I omitted some computation | ||
+ | |||
+ | ~ [[User:Afly|Afly]] ([[User talk:Afly|talk]]) | ||
+ | |||
+ | ==See Also== | ||
+ | {{AIME box|year=2019|n=II|before=First Problem|num-a=2}} | ||
+ | [[Category: Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 14:02, 7 September 2024
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
Solution 6
Let . Now consider
, and if we find the coordinates of
, by symmetry about
, we can find the coordinates of D.
So let . So the following equations hold:
.
.
Solving by squaring both equations and then subtracting one from the other to eliminate , we get
because
is in the second quadrant.
Now by symmetry, .
So now you can proceed by finding the intersection and then calculating the area directly. We get .
~hastapasta
Solution 7
Since the figure formed by connecting the vertices of the congruent triangles is a isoceles trapezoid, by Ptolemys, the other base of the trapezoid is Then, dropping altitudes to the base of
and using pythagorean theorem, we have the height is
and we can use similar triangles to finish.
Solution 8 (Very, very, quick, but for observant people only)
First, let's define H as the intersection of CB and DA, and define G as the midpoint of AB. Next, let E and F be the feet of the perpendicular lines from C and D to line AB respectively. We get a pleasing line of symmetry HG where A maps to B, C maps to D, and E maps to F. We notice that 8-15-17 and 8-6-10 are both pythagorean triples and we test that theory. Then, we find AB = 15 - 6 = 9 and GB = 4.5. Since △CEB~△HGB, we find that HG = 2.4. Then, the area of △HGB is 5.4 and the total overlap is 10.8 = 54/5. Noting that GCD(54,5)=1, we add them to get 59.
Note: I omitted some computation
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.