Difference between revisions of "2021 AIME II Problems/Problem 12"
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==Problem== | ==Problem== | ||
A convex quadrilateral has area <math>30</math> and side lengths <math>5, 6, 9,</math> and <math>7,</math> in that order. Denote by <math>\theta</math> the measure of the acute angle formed by the diagonals of the quadrilateral. Then <math>\tan \theta</math> can be written in the form <math>\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. | A convex quadrilateral has area <math>30</math> and side lengths <math>5, 6, 9,</math> and <math>7,</math> in that order. Denote by <math>\theta</math> the measure of the acute angle formed by the diagonals of the quadrilateral. Then <math>\tan \theta</math> can be written in the form <math>\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. | ||
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==Solution 1 (Sines and Cosines)== | ==Solution 1 (Sines and Cosines)== | ||
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Since we are asked to find <math>\tan \theta</math>, we can find <math>\sin \theta</math> and <math>\cos \theta</math> separately and use their values to get <math>\tan \theta</math>. We can start by drawing a diagram. Let the vertices of the quadrilateral be <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math>. Let <math>AB = 5</math>, <math>BC = 6</math>, <math>CD = 9</math>, and <math>DA = 7</math>. Let <math>AX = a</math>, <math>BX = b</math>, <math>CX = c</math>, and <math>DX = d</math>. We know that <math>\theta</math> is the acute angle formed between the intersection of the diagonals <math>AC</math> and <math>BD</math>. | Since we are asked to find <math>\tan \theta</math>, we can find <math>\sin \theta</math> and <math>\cos \theta</math> separately and use their values to get <math>\tan \theta</math>. We can start by drawing a diagram. Let the vertices of the quadrilateral be <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math>. Let <math>AB = 5</math>, <math>BC = 6</math>, <math>CD = 9</math>, and <math>DA = 7</math>. Let <math>AX = a</math>, <math>BX = b</math>, <math>CX = c</math>, and <math>DX = d</math>. We know that <math>\theta</math> is the acute angle formed between the intersection of the diagonals <math>AC</math> and <math>BD</math>. | ||
<asy> | <asy> | ||
Line 98: | Line 51: | ||
7^2 &= d^2 + a^2 - 2da\cos \theta. | 7^2 &= d^2 + a^2 - 2da\cos \theta. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | We know that <math>\cos (180^\circ - \theta) = -\cos \theta</math>. We can substitute this value into our equations to get: | + | We know that <math>\cos (180^\circ - \theta) = -\cos \theta</math> for all <math>\theta</math>. We can substitute this value into our equations to get: |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | 5^2 &= a^2 + b^2 + 2ab\cos \theta, \\ | + | 5^2 &= a^2 + b^2 + 2ab\cos \theta, &&(1) \\ |
− | 6^2 &= b^2 + c^2 - 2bc\cos \theta, \\ | + | 6^2 &= b^2 + c^2 - 2bc\cos \theta, &&(2) \\ |
− | 9^2 &= c^2 + d^2 + 2cd\cos \theta, \\ | + | 9^2 &= c^2 + d^2 + 2cd\cos \theta, &&(3) \\ |
− | 7^2 &= d^2 + a^2 - 2da\cos \theta. | + | 7^2 &= d^2 + a^2 - 2da\cos \theta. &&(4) |
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | If we subtract | + | If we subtract <math>(2)+(4)</math> from <math>(1)+(3)</math>, the squared terms cancel, leaving us with: |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
5^2 + 9^2 - 6^2 - 7^2 &= 2ab \cos \theta + 2bc \cos \theta + 2cd \cos \theta + 2da \cos \theta \\ | 5^2 + 9^2 - 6^2 - 7^2 &= 2ab \cos \theta + 2bc \cos \theta + 2cd \cos \theta + 2da \cos \theta \\ | ||
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Since we have figured out <math>\sin \theta</math> and <math>\cos \theta</math>, we can calculate <math>\tan \theta</math>: <cmath>\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{60}{ab + bc + cd + da}}{\frac{21/2}{ab + bc + cd + da}} = \frac{60}{21/2} = \frac{120}{21} = \frac{40}{7}.</cmath> | Since we have figured out <math>\sin \theta</math> and <math>\cos \theta</math>, we can calculate <math>\tan \theta</math>: <cmath>\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{60}{ab + bc + cd + da}}{\frac{21/2}{ab + bc + cd + da}} = \frac{60}{21/2} = \frac{120}{21} = \frac{40}{7}.</cmath> | ||
Therefore our answer is <math>40 + 7 = \boxed{047}</math>. | Therefore our answer is <math>40 + 7 = \boxed{047}</math>. | ||
+ | |||
+ | ~ Steven Chen (www.professorchenedu.com) | ||
~ my_aops_lessons | ~ my_aops_lessons | ||
==Solution 2 (Right Triangles)== | ==Solution 2 (Right Triangles)== | ||
− | + | In convex quadrilateral <math>ABCD,</math> let <math>AB=5,BC=6,CD=9,</math> and <math>DA=7.</math> Let <math>A'</math> and <math>C'</math> be the feet of the perpendiculars from <math>A</math> and <math>C,</math> respectively, to <math>\overline{BD}.</math> We obtain the following diagram: | |
+ | <asy> | ||
+ | /* Made by MRENTHUSIASM */ | ||
+ | size(500); | ||
− | + | pair A, B, C, D, P, A1, C1; | |
− | + | B = origin; | |
+ | D = (3*sqrt(32498*(29400*sqrt(47)+312523))/32498,0); | ||
+ | A = intersectionpoints(Circle(B,5),Circle(D,7))[0]; | ||
+ | C = intersectionpoints(Circle(B,6),Circle(D,9))[1]; | ||
+ | P = intersectionpoint(A--C,B--D); | ||
+ | A1 = foot(A,B,D); | ||
+ | C1 = foot(C,B,D); | ||
+ | markscalefactor=3/160; | ||
+ | draw(rightanglemark(A,A1,D),red); | ||
+ | draw(rightanglemark(C,C1,B),red); | ||
+ | dot("$A$",A,1.5*dir(aCos(7/sqrt(1649)))); | ||
+ | dot("$B$",B,1.5*W); | ||
+ | dot("$C$",C,1.5*dir(180+aCos(7/sqrt(1649)))); | ||
+ | dot("$D$",D,1.5*E); | ||
+ | dot("$E$",P,dir(180-(180-aCos(7/sqrt(1649)))/2)); | ||
+ | dot("$A'$",A1,dir(-75)); | ||
+ | dot("$C'$",C1,N); | ||
+ | label("$\theta$",P,dir(180+aCos(7/sqrt(1649))/2),red); | ||
+ | draw(A--A1^^C--C1,dashed); | ||
+ | draw(A--B--C--D--cycle^^A--C^^B--D); | ||
+ | </asy> | ||
Let <math>BC'=p,C'E=q,EA'=r,A'D=s,AA'=h_1,</math> and <math>CC'=h_2.</math> We apply the Pythagorean Theorem to right triangles <math>\triangle ABA',\triangle BCC',\triangle CDC',</math> and <math>\triangle DAA',</math> respectively: | Let <math>BC'=p,C'E=q,EA'=r,A'D=s,AA'=h_1,</math> and <math>CC'=h_2.</math> We apply the Pythagorean Theorem to right triangles <math>\triangle ABA',\triangle BCC',\triangle CDC',</math> and <math>\triangle DAA',</math> respectively: | ||
<cmath>\begin{array}{ccccccccccccccccc} | <cmath>\begin{array}{ccccccccccccccccc} | ||
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<cmath>\begin{align*} | <cmath>\begin{align*} | ||
\frac{h_1+h_2}{q+r}&=\frac{40}{7} \\ | \frac{h_1+h_2}{q+r}&=\frac{40}{7} \\ | ||
− | \frac | + | \frac{r\tan\theta+q\tan\theta}{q+r}&=\frac{40}{7} \hspace{15mm} &&\text{by }(1\star)\text{ and }(2\star)\\ |
\frac{(r+q)\tan\theta}{q+r}&=\frac{40}{7} \\ | \frac{(r+q)\tan\theta}{q+r}&=\frac{40}{7} \\ | ||
\tan\theta&=\frac{40}{7}, | \tan\theta&=\frac{40}{7}, | ||
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | ==Solution 3== | + | ==Solution 3 (Bretschneider's Formula)== |
− | + | ||
+ | ===Bretschneider's Formula=== | ||
+ | |||
+ | <asy> size(200); import olympiad; defaultpen(linewidth(0.8)+fontsize(10)); | ||
+ | pair A,B,C,D; | ||
+ | A=origin; B=(1.25,0); D=dir(65); C=D+0.85*dir(90)*(A-D); | ||
+ | draw(A--B--C--D--cycle); draw(A--C^^B--D, gray+0.5); | ||
+ | dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, NE); dot("$D$",D,NW); | ||
+ | label("$a$", A--B, S); label("$b$", B--C, E); label("$c$", D--C, N); label("$d$",D--A,W); label("$u$",D--B,2*dir(170)); label("$v$",A--C,S); | ||
+ | </asy> | ||
+ | Given quadrilateral <math>ABCD</math>, let, <math>a, b, c, d</math>, be the sides, <math>s</math> the semiperimeter, and <math>u, v</math>, the diagonals. Then the area, <math>K</math>, is given by <cmath>K = \tfrac 14 \sqrt {4u^2v^2-(b^2+d^2-a^2-c^2)^2}</cmath> | ||
+ | |||
+ | ===Solution=== | ||
+ | By Bretschneider's Formula, <cmath>30=\tfrac{1}{4}\sqrt{4u^2v^2-(b^2+d^2-a^2-c^2)^2}=\tfrac{1}{4}\sqrt{4u^2v^2-441}.</cmath> Thus, <math>uv=3\sqrt{1649}</math>. Also, <cmath>[ABCD]=\tfrac 12 \cdot uv\sin{\theta};</cmath> solving for <math>\sin{\theta}</math> yields <math>\sin{\theta}=\tfrac{40}{\sqrt{1649}}</math>. Since <math>\theta</math> is acute, <math>\cos{\theta}</math> is positive, from which <math>\cos{\theta}=\tfrac{7}{\sqrt{1649}}</math>. Solving for <math>\tan{\theta}</math> yields <cmath>\tan{\theta}=\frac{\sin{\theta}}{\cos{\theta}}=\frac{40}{7},</cmath> for a final answer of <math>\boxed{047}</math>. | ||
+ | |||
~ Leo.Euler | ~ Leo.Euler | ||
+ | |||
+ | ==Solution 4 (Symmetry)== | ||
+ | [[File:AIME-I-2021-12.png|350px|right]] | ||
+ | <i><b>Claim</b></i> | ||
+ | |||
+ | Given an inscribed quadrilateral <math>ABCD</math> with sides <math>AB = a, BC = b, CD = c,</math> and <math>DA = d.</math> Prove that the <math>\angle \theta < 90^\circ</math> between the diagonals is given by | ||
+ | <cmath>\begin{align*}2(ac + bd) \cos \theta = {|d^2 – c^2 + b^2 – a^2|}.\end{align*}</cmath> | ||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let the point <math>B'</math> be symmetric to <math>B</math> with respect to the perpendicular bisector <math>AC.</math> Then the quadrilateral <math>AB'CD</math> is an inscribed one, <math>AB' = b, B'C = a.</math> | ||
+ | |||
+ | <cmath> 2 \angle AEB = \overset{\Large\frown} {AB} + \overset{\Large\frown} {CD}.</cmath> | ||
+ | <cmath>\begin{align*} 2\angle B'AD = \overset{\Large\frown} {B'C} + \overset{\Large\frown} {CD} = \overset{\Large\frown} {AB} + \overset{\Large\frown} {CD} \implies \angle AEB = \angle B'AD.\end{align*}</cmath> | ||
+ | |||
+ | We apply the Law of Cosines to <math>\triangle AB'D</math> and <math>\triangle CB'D</math>: | ||
+ | <cmath>\begin{align*} B'D^2 = AD^2 + AB'^2 – 2 AD \cdot AB' \cos \theta, \end{align*}</cmath> | ||
+ | <cmath>\begin{align*} B'D^2 = CD^2 + CB'^2 + 2 CD \cdot CB' \cos \theta,\end{align*}</cmath> | ||
+ | <cmath>\begin{align*} d^2 + b^2 – 2 bd \cos \theta = c^2 + a^2 + 2ac \cos \theta,\end{align*}</cmath> | ||
+ | <cmath>\begin{align*} 2(ac + bd) \cos \theta = |d^2 – c^2 + b^2 – a^2|.\end{align*}</cmath> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Note 1== | ||
+ | By generalization, the tangent of the acute angle formed by the diagonals is <cmath>\left|\frac{4A}{a^2-b^2+c^2-d^2}\right|.</cmath> | ||
+ | |||
+ | ==Video Solution by MOP 2024== | ||
+ | https://youtu.be/K0u0ACMTSw8 | ||
+ | |||
+ | ~r00tsOfUnity | ||
==Video Solution== | ==Video Solution== | ||
https://www.youtube.com/watch?v=7DxIdTLNbo0 | https://www.youtube.com/watch?v=7DxIdTLNbo0 | ||
+ | |||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/8GRO4za5rPI | ||
+ | |||
+ | ~Interstigation | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2021|n=II|num-b=11|num-a=13}} | {{AIME box|year=2021|n=II|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 16:33, 12 January 2024
Contents
Problem
A convex quadrilateral has area and side lengths and in that order. Denote by the measure of the acute angle formed by the diagonals of the quadrilateral. Then can be written in the form , where and are relatively prime positive integers. Find .
Solution 1 (Sines and Cosines)
Since we are asked to find , we can find and separately and use their values to get . We can start by drawing a diagram. Let the vertices of the quadrilateral be , , , and . Let , , , and . Let , , , and . We know that is the acute angle formed between the intersection of the diagonals and . We are given that the area of quadrilateral is . We can express this area using the areas of triangles , , , and . Since we want to find and , we can represent these areas using as follows: We know that . Therefore it follows that: From here we see that . Now we need to find . Using the Law of Cosines on each of the four smaller triangles, we get following equations: We know that for all . We can substitute this value into our equations to get: If we subtract from , the squared terms cancel, leaving us with: From here we see that .
Since we have figured out and , we can calculate : Therefore our answer is .
~ Steven Chen (www.professorchenedu.com)
~ my_aops_lessons
Solution 2 (Right Triangles)
In convex quadrilateral let and Let and be the feet of the perpendiculars from and respectively, to We obtain the following diagram: Let and We apply the Pythagorean Theorem to right triangles and respectively: Let the brackets denote areas. We get We subtract from From right triangles and we have It follows that Finally, we divide by from which the answer is
~MRENTHUSIASM
Solution 3 (Bretschneider's Formula)
Bretschneider's Formula
Given quadrilateral , let, , be the sides, the semiperimeter, and , the diagonals. Then the area, , is given by
Solution
By Bretschneider's Formula, Thus, . Also, solving for yields . Since is acute, is positive, from which . Solving for yields for a final answer of .
~ Leo.Euler
Solution 4 (Symmetry)
Claim
Given an inscribed quadrilateral with sides and Prove that the between the diagonals is given by Proof
Let the point be symmetric to with respect to the perpendicular bisector Then the quadrilateral is an inscribed one,
We apply the Law of Cosines to and :
vladimir.shelomovskii@gmail.com, vvsss
Note 1
By generalization, the tangent of the acute angle formed by the diagonals is
Video Solution by MOP 2024
~r00tsOfUnity
Video Solution
https://www.youtube.com/watch?v=7DxIdTLNbo0
Video Solution by Interstigation
~Interstigation
See Also
2021 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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.