Difference between revisions of "2014 AIME I Problems/Problem 13"
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label("$z$", intersectionpoint( D--P, G--H ));</asy> | label("$z$", intersectionpoint( D--P, G--H ));</asy> | ||
− | == Solution == | + | == Solution (Official Solution, MAA)== |
− | + | Let <math>s</math> be the side length of <math>ABCD</math>, let <math>Q</math>, and <math>R</math> be the midpoints of <math>\overline{EG}</math> and <math>\overline{FH}</math>, respectively, let <math>S</math> be the foot of the perpendicular from <math>Q</math> to <math>\overline{CD}</math>, let <math>T</math> be the foot of the perpendicular from <math>R</math> to <math>\overline{AD}</math>. | |
+ | <asy> size(150); defaultpen(fontsize(10pt)); pair A,B,C,D,E,F,Fp,G,Gp,H,O,I,J,R,S,T; A=dir(45*3); B=dir(-45*3); C=dir(-45); D=dir(45); O = origin; real theta=15; E=extension(A,B,O,dir(180+theta)); G=extension(C,D,O,dir(theta)); I=extension(A,D,O,dir(90+theta)); J=extension(B,C,O,dir(-90+theta)); H=(A+I)/2; F=H+(J-I); R=midpoint(H--F); S=midpoint(C--D); T=(R.x, A.y); draw(A--B--C--D--cycle^^E--G^^F--H, black+0.8); draw(S--R--T, gray+0.4); dotfactor = 3; dot("$A$",A,dir(135)); dot("$B$",B,dir(215)); dot("$C$",C,dir(305)); dot("$D$",D,dir(45)); dot("$H$",H,dir(90)); dot("$F$",F,dir(270)); dot("$E$",E,dir(180)); dot("$G$",G,dir(0)); dot("$Q$",O,dir(-90)); dot("$R$",R,dir(-180)); dot("$S$",S,dir(0)); dot("$T$",T,dir(90)); pair P = extension(F,H,E,G); dot("$P$",P,dir(180+60)); </asy> | ||
+ | The fraction of the area of the square <math>ABCD</math> which is occupied by trapezoid <math>BCGE</math> is <cmath>\frac{275+405}{269+275+405+411}=\frac 12,</cmath>so <math>Q</math> is the center of <math>ABCD</math>. Thus <math>R</math>, <math>Q</math>, <math>S</math> are collinear, and <math>RT=QS=\tfrac 12 s</math>. Similarly, the fraction of the area occupied by trapezoid <math>CDHF</math> is <math>\tfrac 35</math>, so <math>RS=\tfrac 35s</math> and <math>RQ=\tfrac{1}{10}s</math>. | ||
− | + | Because <math>\triangle QSG \cong \triangle RTH</math>, the area of <math>DHPG</math> is the sum <cmath>[DHPG]=[DTRS]+[RPQ].</cmath> Rectangle <math>DTRS</math> has area <math>RS\cdot RT = \tfrac 35s\cdot \tfrac 12 s = \tfrac{3}{10}s^2</math>. If <math>\angle QRP = \theta</math> , then <math>\triangle RPQ</math> has area <cmath>[RPQ]= \tfrac 12 \cdot \tfrac 1{10}s\sin\theta \cdot \tfrac 1{10}s\cos\theta = \tfrac 1{400}s^2\sin 2\theta.</cmath>Therefore the area of <math>[DHPG]</math> is <math>s^2(\tfrac 3{10}+\tfrac 1{400}\sin 2\theta)</math>. Because the area of trapezoid <math>CDHF</math> is <math>\tfrac 35 s^2</math>, the area of <math>CGPF</math> is <math>s^2(\tfrac 3{10}-\tfrac 1{400}\sin 2\theta)</math>. | |
− | + | Because these areas are in the ratio <math>411:405=(408+3):(408-3)</math>, it follows that <cmath>\frac{\frac 1{400}\sin 2\theta}{\frac 3{10}}=\frac 3{408},</cmath>from which we get <math>\sin 2\theta = \tfrac {15}{17}</math>. Note that <math>\theta =\angle RHT > \angle QAT = 45^\circ</math>, so <math>\cos 2\theta = -\sqrt{1-\sin^2 2\theta}= -\tfrac 8{17}</math> and <math>\sin^2\theta = \tfrac{1}{2}(1-\cos 2\theta) = \tfrac{25}{34}</math>. Then <cmath>[ABCD]=s^2 = EG^2\sin^2\theta = 34^2 \cdot \tfrac {25}{34} = 850.</cmath> | |
− | + | == Solution 1== | |
− | Draw <math>\overline{ | + | Let <math>s</math> be the side length of <math>ABCD</math>, let <math>[ABCD]=1360a</math>. Let <math>Q</math> and <math>R</math> be the midpoints of <math>\overline{EG}</math> and <math>\overline{FH}</math>, respectively; because <math>269+411=275+405</math>, <math>Q</math> is also the center of the square. Draw <math>\overline{IJ} \parallel \overline{HF}</math> through <math>Q</math>, with <math>I</math> on <math>\overline{AD}</math>, <math>J</math> on <math>\overline{BC}</math>. |
+ | <asy> size(150); | ||
+ | defaultpen(fontsize(9pt)); | ||
+ | pair A,B,C,D,E,F,G,H,I,J,L,P,Q,R,S; | ||
+ | Q=MP("Q",origin,down); A=MP("A",(-1,1),dir(135)); B=MP("B",(-1,-1),dir(225)); C=MP("C",(1,-1),dir(-45)); D=MP("D",(1,1),dir(45)); real theta = 20; real shift=0.4; E=MP("E",extension(A,B,Q,dir(theta)),left); J=MP("J",extension(B,C,Q,dir(90+theta)),down); F=MP("F",J+(shift*left),down); G=MP("G",extension(C,D,Q,dir(theta)),right); I=MP("I",extension(A,D,Q,dir(90+theta)),up); H=MP("H",I+(shift*left),up); P=MP("P",extension(E,G,F,H),2*dir(-110)); R=MP("R",extension(F,H,Q,left),left); | ||
+ | draw(A--B--C--D--cycle^^E--G^^F--H, black+1); draw(R--Q^^I--J, gray); | ||
+ | </asy> | ||
+ | Segments <math>\overline{EG}</math> and <math>\overline{IJ}</math> divide the square into four congruent quadrilaterals, each of area <math>\tfrac 14 [ABCD]=340a</math>. Then <cmath>[HFJI]=[ABJI]-[ABFH]=136a.</cmath> The fraction of the total area occupied by parallelogram <math>HFJI</math> is <math>\tfrac 1{10}</math>, so <math>RQ=\tfrac{s}{10}</math>. | ||
− | + | Because <math>[HFJI]= HF\cdot PQ</math>, with <math>HF=34</math>, we get <math>PQ=4a</math>. | |
+ | Now <cmath>[PQR]=[HPQI]-[HRQI]= ([AEQI]-[AEPH])-\tfrac 12[IJFH] = 3a,</cmath> and because <math>[PQR]=\tfrac 12 \cdot PQ\cdot PR</math>, with <math>PQ=4a</math>, we get <math>PR=\tfrac 32</math>. | ||
+ | By Pythagoras' Theorem on <math>\triangle PQR</math>, we get <cmath>16a^2+\frac 94 =\tfrac{68}{5}a,\quad \text{i.e.}\quad 320a^2-272a+45=0,</cmath> with roots <math>a=\tfrac 9{40}</math> or <math>a=\tfrac58</math>. The former leads to a square with diagonal less than <math>34</math>, which can't be, since <math>EG=FH=34</math>; therefore <math>a=\tfrac 58</math> and <math>[ABCD]=850</math>. | ||
− | + | ==Solution 2 (Fakesolve)== | |
− | + | <math>269+275+405+411=1360</math>, a multiple of <math>17</math>. In addition, <math>EG=FH=34</math>, which is <math>17\cdot 2</math>. | |
− | + | Therefore, we suspect the square of the "hypotenuse" of a right triangle, corresponding to <math>EG</math> and <math>FH</math> must be a multiple of <math>17</math>. All of these triples are primitive: | |
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− | Therefore the | ||
<cmath>17=1^2+4^2</cmath> | <cmath>17=1^2+4^2</cmath> | ||
<cmath>34=3^2+5^2</cmath> | <cmath>34=3^2+5^2</cmath> | ||
<cmath>51=\emptyset</cmath> | <cmath>51=\emptyset</cmath> | ||
+ | <cmath>68=\emptyset\text{ others}</cmath> | ||
<cmath>85=2^2+9^2=6^2+7^2</cmath> | <cmath>85=2^2+9^2=6^2+7^2</cmath> | ||
<cmath>102=\emptyset</cmath> | <cmath>102=\emptyset</cmath> | ||
<cmath>119=\emptyset \dots</cmath> | <cmath>119=\emptyset \dots</cmath> | ||
− | The sides of the square | + | The sides of the square can only equal the longer leg, or else the lines would have to extend outside of the square. Substituting <math>EG=FH=34</math>: |
− | <cmath>\sqrt{17}\rightarrow 34\implies | + | <cmath>\sqrt{17}\rightarrow 34\implies 8\sqrt{17}\implies A=\textcolor{red}{1088}</cmath> |
− | <cmath>\sqrt{34}\rightarrow 34\implies | + | <cmath>\sqrt{34}\rightarrow 34\implies 5\sqrt{34}\implies A=850</cmath> |
− | <cmath>\sqrt{85}\rightarrow 34\implies \{ | + | <cmath>\sqrt{85}\rightarrow 34\implies \{18\sqrt{85}/5,14\sqrt{85}/5\}\implies A=\textcolor{red}{1101.6,666.4}</cmath> |
+ | |||
+ | Thus, <math>\boxed{850}</math> is the only valid answer. | ||
+ | |||
+ | == Solution 3== | ||
+ | |||
+ | Continue in the same way as solution 1 to get that <math>POK</math> has area <math>3a</math>, and <math>OK = \frac{d}{10}</math>. You can then find <math>PK</math> has length <math>\frac 32</math>. | ||
+ | |||
+ | Then, if we drop a perpendicular from <math>H</math> to <math>BC</math> at <math>L</math>, We get <math>\triangle HLF \sim \triangle OPK</math>. | ||
+ | |||
+ | Thus, <math>LF = \frac{15\cdot 34}{d}</math>, and we know <math>HL = d</math>, and <math>HF = 34</math>. Thus, we can set up an equation in terms of <math>d</math> using the Pythagorean theorem. | ||
+ | |||
+ | <cmath>\frac{15^2 \cdot 34^2}{d^2} + d^2 = 34^2</cmath> | ||
+ | |||
+ | <cmath>d^4 - 34^2 d^2 + 15^2 \cdot 34^2 = 0</cmath> | ||
+ | |||
+ | <cmath>(d^2 - 34 \cdot 25)(d^2 - 34 \cdot 9) = 0</cmath> | ||
+ | |||
+ | <math>d^2 = 34 \cdot 9</math> is extraneous, so <math>d^2 = 34 \cdot 25</math>. Since the area is <math>d^2</math>, we have it is equal to <math>34 \cdot 25 = \boxed{850}</math> | ||
+ | |||
+ | -Alexlikemath | ||
+ | |||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/Kcug2ALOjkA?si=VoImhnX5rAKhprgk | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
+ | |||
+ | |||
+ | |||
+ | ==Video Solution by Punxsutawney Phil== | ||
− | + | https://youtube.com/watch?v=wrxET2c0ZgU | |
== See also == | == See also == | ||
{{AIME box|year=2014|n=I|num-b=12|num-a=14}} | {{AIME box|year=2014|n=I|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:07, 13 October 2023
Contents
Problem 13
On square , points , and lie on sides and respectively, so that and . Segments and intersect at a point , and the areas of the quadrilaterals and are in the ratio Find the area of square .
Solution (Official Solution, MAA)
Let be the side length of , let , and be the midpoints of and , respectively, let be the foot of the perpendicular from to , let be the foot of the perpendicular from to . The fraction of the area of the square which is occupied by trapezoid is so is the center of . Thus , , are collinear, and . Similarly, the fraction of the area occupied by trapezoid is , so and .
Because , the area of is the sum Rectangle has area . If , then has area Therefore the area of is . Because the area of trapezoid is , the area of is .
Because these areas are in the ratio , it follows that from which we get . Note that , so and . Then
Solution 1
Let be the side length of , let . Let and be the midpoints of and , respectively; because , is also the center of the square. Draw through , with on , on . Segments and divide the square into four congruent quadrilaterals, each of area . Then The fraction of the total area occupied by parallelogram is , so .
Because , with , we get . Now and because , with , we get . By Pythagoras' Theorem on , we get with roots or . The former leads to a square with diagonal less than , which can't be, since ; therefore and .
Solution 2 (Fakesolve)
, a multiple of . In addition, , which is . Therefore, we suspect the square of the "hypotenuse" of a right triangle, corresponding to and must be a multiple of . All of these triples are primitive:
The sides of the square can only equal the longer leg, or else the lines would have to extend outside of the square. Substituting :
Thus, is the only valid answer.
Solution 3
Continue in the same way as solution 1 to get that has area , and . You can then find has length .
Then, if we drop a perpendicular from to at , We get .
Thus, , and we know , and . Thus, we can set up an equation in terms of using the Pythagorean theorem.
is extraneous, so . Since the area is , we have it is equal to
-Alexlikemath
Video Solution
https://youtu.be/Kcug2ALOjkA?si=VoImhnX5rAKhprgk
~MathProblemSolvingSkills.com
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=wrxET2c0ZgU
See also
2014 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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