Difference between revisions of "2021 AIME II Problems/Problem 12"
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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>. | ||
− | + | ==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>. | ||
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<asy> | <asy> | ||
unitsize(4cm); | unitsize(4cm); | ||
Line 153: | Line 31: | ||
label("$d$",(D+X)/2,SE); | label("$d$",(D+X)/2,SE); | ||
</asy> | </asy> | ||
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We are given that the area of quadrilateral <math>ABCD</math> is <math>30</math>. We can express this area using the areas of triangles <math>AXB</math>, <math>BXC</math>, <math>CXD</math>, and <math>DXA</math>. Since we want to find <math>\sin \theta</math> and <math>\cos \theta</math>, we can represent these areas using <math>\sin \theta</math> as follows: | We are given that the area of quadrilateral <math>ABCD</math> is <math>30</math>. We can express this area using the areas of triangles <math>AXB</math>, <math>BXC</math>, <math>CXD</math>, and <math>DXA</math>. Since we want to find <math>\sin \theta</math> and <math>\cos \theta</math>, we can represent these areas using <math>\sin \theta</math> as follows: | ||
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<cmath>\begin{align*} | <cmath>\begin{align*} | ||
30 &=[ABCD] \\ | 30 &=[ABCD] \\ | ||
&=[AXB] + [BXC] + [CXD] + [DXA] \\ | &=[AXB] + [BXC] + [CXD] + [DXA] \\ | ||
&=\frac{1}{2} ab \sin (\angle AXB) + \frac{1}{2} bc \sin (\angle BXC) + \frac{1}{2} cd \sin (\angle CXD) + \frac{1}{2} da \sin (\angle AXD) \\ | &=\frac{1}{2} ab \sin (\angle AXB) + \frac{1}{2} bc \sin (\angle BXC) + \frac{1}{2} cd \sin (\angle CXD) + \frac{1}{2} da \sin (\angle AXD) \\ | ||
− | &=\frac{1}{2} ab \sin (180^\circ - \theta) + \frac{1}{2} bc \sin (\theta) + \frac{1}{2} cd \sin (180^\circ - \theta) + \frac{1}{2} da \sin (\theta) | + | &=\frac{1}{2} ab \sin (180^\circ - \theta) + \frac{1}{2} bc \sin (\theta) + \frac{1}{2} cd \sin (180^\circ - \theta) + \frac{1}{2} da \sin (\theta). |
\end{align*}</cmath> | \end{align*}</cmath> | ||
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We know that <math>\sin (180^\circ - \theta) = \sin \theta</math>. Therefore it follows that: | We know that <math>\sin (180^\circ - \theta) = \sin \theta</math>. Therefore it follows that: | ||
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<cmath>\begin{align*} | <cmath>\begin{align*} | ||
30 &=\frac{1}{2} ab \sin (180^\circ - \theta) + \frac{1}{2} bc \sin (\theta) + \frac{1}{2} cd \sin (180^\circ - \theta) + \frac{1}{2} da \sin (\theta) \\ | 30 &=\frac{1}{2} ab \sin (180^\circ - \theta) + \frac{1}{2} bc \sin (\theta) + \frac{1}{2} cd \sin (180^\circ - \theta) + \frac{1}{2} da \sin (\theta) \\ | ||
&=\frac{1}{2} ab \sin (\theta) + \frac{1}{2} bc \sin (\theta) + \frac{1}{2} cd \sin (\theta) + \frac{1}{2} da \sin (\theta) \\ | &=\frac{1}{2} ab \sin (\theta) + \frac{1}{2} bc \sin (\theta) + \frac{1}{2} cd \sin (\theta) + \frac{1}{2} da \sin (\theta) \\ | ||
− | &=\frac{1}{2}\sin\theta (ab + bc + cd + da) | + | &=\frac{1}{2}\sin\theta (ab + bc + cd + da). |
\end{align*}</cmath> | \end{align*}</cmath> | ||
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From here we see that <math>\sin \theta = \frac{60}{ab + bc + cd + da}</math>. Now we need to find <math>\cos \theta</math>. Using the Law of Cosines on each of the four smaller triangles, we get following equations: | From here we see that <math>\sin \theta = \frac{60}{ab + bc + cd + da}</math>. Now we need to find <math>\cos \theta</math>. Using the Law of Cosines on each of the four smaller triangles, we get following equations: | ||
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<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | 5^2 &= a^2 + b^2 - 2ab\cos(180^\circ-\theta) \\ | + | 5^2 &= a^2 + b^2 - 2ab\cos(180^\circ-\theta), \\ |
− | 6^2 &= b^2 + c^2 - 2bc\cos \theta \\ | + | 6^2 &= b^2 + c^2 - 2bc\cos \theta, \\ |
− | 9^2 &= c^2 + d^2 - 2cd\cos(180^\circ-\theta) \\ | + | 9^2 &= c^2 + d^2 - 2cd\cos(180^\circ-\theta), \\ |
− | 7^2 &= d^2 + a^2 - 2da\cos \theta | + | 7^2 &= d^2 + a^2 - 2da\cos \theta. |
+ | \end{align*}</cmath> | ||
+ | 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*} | ||
+ | 5^2 &= a^2 + b^2 + 2ab\cos \theta, &&(1) \\ | ||
+ | 6^2 &= b^2 + c^2 - 2bc\cos \theta, &&(2) \\ | ||
+ | 9^2 &= c^2 + d^2 + 2cd\cos \theta, &&(3) \\ | ||
+ | 7^2 &= d^2 + a^2 - 2da\cos \theta. &&(4) | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
+ | If we subtract <math>(2)+(4)</math> from <math>(1)+(3)</math>, the squared terms cancel, leaving us with: | ||
+ | <cmath>\begin{align*} | ||
+ | 5^2 + 9^2 - 6^2 - 7^2 &= 2ab \cos \theta + 2bc \cos \theta + 2cd \cos \theta + 2da \cos \theta \\ | ||
+ | 21 &= 2\cos \theta (ab + bc + cd + da). | ||
+ | \end{align*}</cmath> | ||
+ | From here we see that <math>\cos \theta = \frac{21/2}{ab + bc + cd + da}</math>. | ||
− | + | 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>. | ||
− | <cmath>\begin{ | + | ~ Steven Chen (www.professorchenedu.com) |
− | 5^2 &= | + | |
− | + | ~ my_aops_lessons | |
− | + | ||
− | 7 | + | ==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: | ||
+ | <cmath>\begin{array}{ccccccccccccccccc} | ||
+ | (p+q+r)^2&+&h_1^2&=&5^2, &&&&&&&&&&&&\hspace{36mm}(1) \\ [1ex] | ||
+ | p^2&+&h_2^2&=&6^2, &&&&&&&&&&&&\hspace{36mm}(2) \\ [1ex] | ||
+ | (q+r+s)^2&+&h_2^2&=&9^2, &&&&&&&&&&&&\hspace{36mm}(3) \\ [1ex] | ||
+ | s^2&+&h_1^2&=&7^2. &&&&&&&&&&&&\hspace{36mm}(4) | ||
+ | \end{array}</cmath> | ||
+ | Let the brackets denote areas. We get | ||
+ | <cmath>\begin{align*} | ||
+ | [ABD]+[CBD]&=[ABCD] \\ | ||
+ | \frac12(p+q+r+s)h_1+\frac12(p+q+r+s)h_2&=30 \\ | ||
+ | \frac12(p+q+r+s)(h_1+h_2)&=30 \\ | ||
+ | (p+q+r+s)(h_1+h_2)&=60. \hspace{49.25mm}(5) | ||
+ | \end{align*}</cmath> | ||
+ | We subtract <math>(2)+(4)</math> from <math>(1)+(3):</math> | ||
+ | <cmath>\begin{align*} | ||
+ | (p+q+r)^2+(q+r+s)^2-p^2-s^2&=21 \\ | ||
+ | \left[(p+q+r)^2-s^2\right]+\left[(q+r+s)^2-p^2\right]&=21 \\ | ||
+ | (p+q+r+s)(p+q+r-s)+(p+q+r+s)(-p+q+r+s)&=21 \\ | ||
+ | (p+q+r+s)(2q+2r)&=21 \\ | ||
+ | 2(p+q+r+s)(q+r)&=21 \\ | ||
+ | (p+q+r+s)(q+r)&=\frac{21}{2}. \hspace{9.5mm}(6) | ||
+ | \end{align*}</cmath> | ||
+ | From right triangles <math>\triangle AEA'</math> and <math>\triangle CEC',</math> we have <math>\tan\theta=\frac{h_1}{r}=\frac{h_2}{q}.</math> It follows that | ||
+ | <cmath>\begin{alignat*}{8} | ||
+ | \tan\theta&=\frac{h_1}{r}\qquad&\implies\qquad h_1&=r\tan\theta, \hspace{64mm}&(1\star)\\ | ||
+ | \tan\theta&=\frac{h_2}{q}\qquad&\implies\qquad h_2&=q\tan\theta. &(2\star) | ||
+ | \end{alignat*}</cmath> | ||
+ | Finally, we divide <math>(5)</math> by <math>(6):</math> | ||
+ | <cmath>\begin{align*} | ||
+ | \frac{h_1+h_2}{q+r}&=\frac{40}{7} \\ | ||
+ | \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} \\ | ||
+ | \tan\theta&=\frac{40}{7}, | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
+ | from which the answer is <math>40+7=\boxed{047}.</math> | ||
+ | |||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | ==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 | |
− | |||
− | |||
− | + | ==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 | ||
− | ==See | + | ==Video Solution by Interstigation== |
+ | https://youtu.be/8GRO4za5rPI | ||
+ | |||
+ | ~Interstigation | ||
+ | |||
+ | ==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 17:33, 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.