Difference between revisions of "2021 AIME I Problems/Problem 11"

m (Remark (Ptolemy's Theorem))
m (Solution 4 (Symmetry): Stated the final answer.)
(29 intermediate revisions by 2 users not shown)
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==Diagram==
 
==Diagram==
[[File:2021 AIME I Problem 11 Diagram.png|center]]
+
<asy>
 +
/* Made by MRENTHUSIASM */
 +
size(300);
  
~MRENTHUSIASM (by Geometry Expressions)
+
pair A, B, C, D, A1, B1, C1, D1;
 
+
A = origin;
==Solution 1==
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C = (sqrt(53041)/31,0);
[[File:Leonard_my_dude's_image.png|center]]
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B = intersectionpoints(Circle(A,4),Circle(C,5))[0];
 
+
D = intersectionpoints(Circle(A,7),Circle(C,6))[1];
Let <math>O</math> be the intersection of <math>AC</math> and <math>BD</math>. Let <math>\theta = \angle AOB</math>.
+
A1 = foot(A,B,D);
 
+
C1 = foot(C,B,D);
Firstly, since <math>\angle AA_1D = \angle AD_1D = 90^\circ</math>, we deduce that <math>AA_1D_1D</math> is cyclic. This implies that <math>\triangle A_1OD_1 \sim \triangle AOD</math>, with a ratio of <math>\frac{A_1O}{AO} = \cos \angle A_1OA = \cos \theta</math>. This means that <math>\frac{A_1D_1}{AD} = \cos \theta</math>. Similarly, <math>\frac{A_1B_1}{AB} = \frac{B_1C_1}{BC} = \frac{C_1D_1}{CD} = \cos \theta</math>. Hence <cmath>A_1B_1 + B_1C_1 + C_1D_1 + D_1A_1 = (AB + BC + CD + DA)\cos \theta</cmath> It therefore only remains to find <math>\cos \theta</math>.
+
B1 = foot(B,A,C);
 
+
D1 = foot(D,A,C);
From Ptolemy's theorem, we have that <math>(BD)(AC) = 4\times6+5\times7 = 59</math>. From Brahmagupta's Formula, <math>[ABCD] = \sqrt{(11-4)(11-5)(11-6)(11-7)} = 2\sqrt{210}</math>. But the area is also <math>\frac{1}{2}(BD)(AC)\sin\theta = \frac{59}{2}\sin\theta</math>, so <math>\sin \theta = \frac{4\sqrt{210}}{59} \implies \cos \theta = \frac{11}{59}</math>. Then the desired fraction is <math>(4+5+6+7)\cos\theta = \frac{242}{59}</math> for an answer of <math>\boxed{301}</math>.
+
markscalefactor=0.025;
 +
draw(rightanglemark(A,A1,B),red);
 +
draw(rightanglemark(B,B1,A),red);
 +
draw(rightanglemark(C,C1,D),red);
 +
draw(rightanglemark(D,D1,C),red);
 +
draw(A1--B1--C1--D1--cycle,green);
 +
dot("$A$",A,1.5*W,linewidth(4));
 +
dot("$B$",B,1.5*dir(180-aCos(11/59)),linewidth(4));
 +
dot("$C$",C,1.5*E,linewidth(4));
 +
dot("$D$",D,1.5*dir(-aCos(11/59)),linewidth(4));
 +
dot("$A_1$",A1,1.5*dir(A1-A),linewidth(4));
 +
dot("$B_1$",B1,1.5*S,linewidth(4));
 +
dot("$C_1$",C1,1.5*dir(C1-C),linewidth(4));
 +
dot("$D_1$",D1,1.5*N,linewidth(4));
 +
draw(A--B--C--D--cycle^^A--C^^B--D^^circumcircle(A,B,C));
 +
draw(A--A1^^B--B1^^C--C1^^D--D1,dashed);
 +
</asy>
 +
~MRENTHUSIASM
  
==Solution 2 (Finding cos x)==
+
==Solution 1 (Cyclic Quadrilaterals, Similar Triangles, and Trigonometry)==
The angle <math>\theta</math> between diagonals satisfies <cmath>\tan{\frac{\theta}{2}}=\sqrt{\frac{(s-b)(s-d)}{(s-a)(s-c)}}</cmath> (see https://en.wikipedia.org/wiki/Cyclic_quadrilateral#Angle_formulas).
 
Thus, <cmath>\tan{\frac{\theta}{2}}=\sqrt{\frac{(11-4)(11-6)}{(11-5)(11-7)}}</cmath> or <cmath>\tan{\frac{\theta}{2}}=\sqrt{\frac{(11-5)(11-7)}{(11-4)(11-6)}}</cmath>
 
That is, <math>\tan^2{\frac{\theta}{2}}=\frac{1-\cos^2{\frac{\theta}{2}}}{\cos^2{\frac{\theta}{2}}}=\frac{24}{35}</math> or <math>\frac{35}{24}</math>
 
Thus, <math>\cos^2{\frac{\theta}{2}}=\frac{35}{59}</math> or <math>\frac{24}{59}</math>
 
<cmath>\cos{\theta}=2\cos^2{\frac{\theta}{2}}-1=\frac{\pm11}{59}</cmath>
 
In this context, <math>\cos{\theta}>0</math>. Thus, <math>\cos{\theta}=\frac{11}{59}</math>
 
<cmath>Ans=22*\cos{\theta}=22*\frac{11}{59}=\frac{242}{59}=\frac{m}{n}</cmath>
 
<cmath>m+n=242+59=\boxed{301}</cmath>
 
~y.grace.yu
 
 
 
==Solution 3 (Pythagorean Theorem)==
 
We assume that the two quadrilateral mentioned in the problem are similar (due to both of them being cyclic). Note that by Ptolemy’s, one of the diagonals has length <math>\sqrt{4 \cdot 6 + 5 \cdot 7} = \sqrt{59}.</math> [I don't believe this is correct... are the two diagonals of <math>ABCD</math> necessarily congruent? -peace09] WLOG we focus on diagonal <math>BD.</math> To find the diagonal of the inner quadrilateral, we drop the altitude from <math>A</math> and <math>C</math> and calculate the length of <math>A_1C_1.</math> Let <math>x</math> be <math>A_1D</math> (Thus <math>A_1B = \sqrt{59} - x.</math> By Pythagorean theorem, we have <cmath>49 - x^2 = 16 - (\sqrt{59} - x)^2 \implies 92 = 2\sqrt{59}x \implies x = \frac{46}{\sqrt{59}} = \frac{46\sqrt{59}}{59}.</cmath> Now let <math>y</math> be <math>C_1D.</math> (thus making <math>C_1B = \sqrt{59} - y</math>). Similarly, we have <cmath>36 - y^2 = 25 - (\sqrt{59} - y)^2 \implies 70 = 2\sqrt{59}y \implies y = \frac{35}{\sqrt{59}} = \frac{35\sqrt{59}}{59}.</cmath> We see that <math>A_1C_1</math>, the scaled down diagonal is just <math>x - y = \frac{11\sqrt{59}}{59},</math> which is <math>\frac{\frac{11\sqrt{59}}{59}}{\sqrt{59}} = \frac{11}{59}</math> times our original diagonal <math>BD,</math> implying a scale factor of <math>\frac{11}{59}.</math> Thus, due to perimeters scaling linearly, the perimeter of the new quadrilateral is simply <math>\frac{11}{59} \cdot 22 = \frac{242}{59},</math> making our answer <math>242+59 = \boxed{301}.</math>
 
-fidgetboss_4000
 
 
 
==Solution 4 (Cyclic Quadrilaterals, Similar Triangles, and Trigonometry)==
 
 
This solution refers to the <b>Diagram</b> section.
 
This solution refers to the <b>Diagram</b> section.
  
Line 39: Line 43:
  
 
We obtain the following diagram:
 
We obtain the following diagram:
[[File:2021 AIME I Problem 11 Solution.png|center]]
+
<asy>
 +
/* Made by MRENTHUSIASM */
 +
size(300);
 +
 
 +
pair A, B, C, D, A1, B1, C1, D1, P, M1, M2;
 +
A = origin;
 +
C = (sqrt(53041)/31,0);
 +
B = intersectionpoints(Circle(A,4),Circle(C,5))[0];
 +
D = intersectionpoints(Circle(A,7),Circle(C,6))[1];
 +
A1 = foot(A,B,D);
 +
C1 = foot(C,B,D);
 +
B1 = foot(B,A,C);
 +
D1 = foot(D,A,C);
 +
P = intersectionpoint(A--C,B--D);
 +
M1 = midpoint(A--B);
 +
M2 = midpoint(C--D);
 +
markscalefactor=0.025;
 +
draw(rightanglemark(A,A1,B),red);
 +
draw(rightanglemark(B,B1,A),red);
 +
draw(rightanglemark(C,C1,D),red);
 +
draw(rightanglemark(D,D1,C),red);
 +
draw(Arc(M1,A,B)^^Arc(M2,C,D),blue);
 +
draw(A1--B1--C1--D1--cycle,green);
 +
dot("$A$",A,1.5*W,linewidth(4));
 +
dot("$B$",B,1.5*dir(180-aCos(11/59)),linewidth(4));
 +
dot("$C$",C,1.5*E,linewidth(4));
 +
dot("$D$",D,1.5*dir(-aCos(11/59)),linewidth(4));
 +
dot("$A_1$",A1,1.5*dir(A1-A),linewidth(4));
 +
dot("$B_1$",B1,1.5*S,linewidth(4));
 +
dot("$C_1$",C1,1.5*dir(C1-C),linewidth(4));
 +
dot("$D_1$",D1,1.5*N,linewidth(4));
 +
dot("$E$",P,dir((180-aCos(11/59))/2),linewidth(4));
 +
label("$\theta$",P,dir(180-aCos(11/59)/2),red);
 +
draw(A--B--C--D--cycle^^A--C^^B--D^^circumcircle(A,B,C));
 +
draw(A--A1^^B--B1^^C--C1^^D--D1,dashed);
 +
</asy>
 
In every cyclic quadrilateral, each pair of opposite angles is supplementary. So, we have <math>\angle EA_1B_1=\angle EAB</math> (both supplementary to <math>\angle B_1A_1B</math>) and <math>\angle EB_1A_1=\angle EBA</math> (both supplementary to <math>\angle A_1B_1A</math>), from which <math>\triangle A_1B_1E \sim \triangle ABE</math> by AA, with the ratio of similitude <cmath>\frac{A_1B_1}{AB}=\underbrace{\frac{A_1E}{AE}}_{\substack{\text{right} \\ \triangle A_1AE}}=\underbrace{\frac{B_1E}{BE}}_{\substack{\text{right} \\ \triangle B_1BE}}=\cos\theta. \hspace{15mm}(1)</cmath>
 
In every cyclic quadrilateral, each pair of opposite angles is supplementary. So, we have <math>\angle EA_1B_1=\angle EAB</math> (both supplementary to <math>\angle B_1A_1B</math>) and <math>\angle EB_1A_1=\angle EBA</math> (both supplementary to <math>\angle A_1B_1A</math>), from which <math>\triangle A_1B_1E \sim \triangle ABE</math> by AA, with the ratio of similitude <cmath>\frac{A_1B_1}{AB}=\underbrace{\frac{A_1E}{AE}}_{\substack{\text{right} \\ \triangle A_1AE}}=\underbrace{\frac{B_1E}{BE}}_{\substack{\text{right} \\ \triangle B_1BE}}=\cos\theta. \hspace{15mm}(1)</cmath>
 
Similarly, we have <math>\angle EC_1D_1=\angle ECD</math> (both supplementary to <math>\angle D_1C_1D</math>) and <math>\angle ED_1C_1=\angle EDC</math> (both supplementary to <math>\angle C_1D_1C</math>), from which <math>\triangle C_1D_1E \sim \triangle CDE</math> by AA, with the ratio of similitude <cmath>\frac{C_1D_1}{CD}=\underbrace{\frac{C_1E}{CE}}_{\substack{\text{right} \\ \triangle C_1CE}}=\underbrace{\frac{D_1E}{DE}}_{\substack{\text{right} \\ \triangle D_1DE}}=\cos\theta. \hspace{14.75mm}(2)</cmath>
 
Similarly, we have <math>\angle EC_1D_1=\angle ECD</math> (both supplementary to <math>\angle D_1C_1D</math>) and <math>\angle ED_1C_1=\angle EDC</math> (both supplementary to <math>\angle C_1D_1C</math>), from which <math>\triangle C_1D_1E \sim \triangle CDE</math> by AA, with the ratio of similitude <cmath>\frac{C_1D_1}{CD}=\underbrace{\frac{C_1E}{CE}}_{\substack{\text{right} \\ \triangle C_1CE}}=\underbrace{\frac{D_1E}{DE}}_{\substack{\text{right} \\ \triangle D_1DE}}=\cos\theta. \hspace{14.75mm}(2)</cmath>
Line 55: Line 94:
 
Two solutions follow from here:
 
Two solutions follow from here:
  
===Solution 4.1 (Law of Cosines)===
+
===Solution 1.1 (Law of Cosines)===
 
Note that <math>\cos(180^\circ-\theta)=-\cos\theta</math> holds for all <math>\theta.</math> We apply the Law of Cosines to <math>\triangle ABE, \triangle BCE, \triangle CDE,</math> and <math>\triangle DAE,</math> respectively:
 
Note that <math>\cos(180^\circ-\theta)=-\cos\theta</math> holds for all <math>\theta.</math> We apply the Law of Cosines to <math>\triangle ABE, \triangle BCE, \triangle CDE,</math> and <math>\triangle DAE,</math> respectively:
 
<cmath>\begin{alignat*}{12}
 
<cmath>\begin{alignat*}{12}
Line 72: Line 111:
 
Finally, substituting this result into <math>(\bigstar)</math> gives <math>22\cos\theta=\frac{242}{59},</math> from which the answer is <math>242+59=\boxed{301}.</math>
 
Finally, substituting this result into <math>(\bigstar)</math> gives <math>22\cos\theta=\frac{242}{59},</math> from which the answer is <math>242+59=\boxed{301}.</math>
  
~MRENTHUSIASM (inspired by Math Jams's <b>2021 AIME I Discussion</b>)
+
~MRENTHUSIASM (credit given to Math Jams's <b>2021 AIME I Discussion</b>)
  
===Solution 4.2 (Area Formulas)===
+
===Solution 1.2 (Area Formulas)===
 
Let the brackets denote areas.  
 
Let the brackets denote areas.  
 
We find <math>[ABCD]</math> in two different ways:
 
We find <math>[ABCD]</math> in two different ways:
Line 92: Line 131:
 
Finally, substituting this result into <math>(\bigstar)</math> gives <math>22\cos\theta=\frac{242}{59},</math> from which the answer is <math>242+59=\boxed{301}.</math>
 
Finally, substituting this result into <math>(\bigstar)</math> gives <math>22\cos\theta=\frac{242}{59},</math> from which the answer is <math>242+59=\boxed{301}.</math>
  
~MRENTHUSIASM
+
~MRENTHUSIASM (credit given to Leonard my dude)
  
 
===Remark (Ptolemy's Theorem)===
 
===Remark (Ptolemy's Theorem)===
 
In <math>ABCD,</math> we have
 
In <math>ABCD,</math> we have
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
AE\cdot BE+BE\cdot CE+CE\cdot DE+DE\cdot AE &= (AE+CE)(BE+DE) &&\hspace{5mm}\text{Factor} \\
+
AE\cdot BE+BE\cdot CE+CE\cdot DE+DE\cdot AE &= (AE+CE)(BE+DE) &&\hspace{10mm}\text{Factor by Grouping} \\
&=AC\cdot BD &&\hspace{5mm}\text{Segment Addition} \\
+
&=AC\cdot BD &&\hspace{10mm}\text{Segment Addition} \\
&=AB\cdot CD+BC\cdot DA &&\hspace{5mm}\text{Ptolemy's Theorem} \\
+
&=AB\cdot CD+BC\cdot DA &&\hspace{10mm}\text{Ptolemy's Theorem} \\
&=59.
+
&=59. &&\hspace{10mm}\text{Substitution}
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
==See also==
+
==Solution 2 (Finding cos x)==
 +
The angle <math>\theta</math> between diagonals satisfies <cmath>\tan{\frac{\theta}{2}}=\sqrt{\frac{(s-b)(s-d)}{(s-a)(s-c)}}</cmath> (see https://en.wikipedia.org/wiki/Cyclic_quadrilateral#Angle_formulas).
 +
Thus, <cmath>\tan{\frac{\theta}{2}}=\sqrt{\frac{(11-4)(11-6)}{(11-5)(11-7)}}\text{ or }\tan{\frac{\theta}{2}}=\sqrt{\frac{(11-5)(11-7)}{(11-4)(11-6)}}.</cmath>
 +
That is, <math>\tan^2{\frac{\theta}{2}}=\frac{1-\cos^2{\frac{\theta}{2}}}{\cos^2{\frac{\theta}{2}}}=\frac{24}{35}</math> or <math>\frac{35}{24}</math>. Thus, <math>\cos^2{\frac{\theta}{2}}=\frac{35}{59}</math> or <math>\frac{24}{59}</math>. So,
 +
<cmath>\cos{\theta}=2\cos^2{\frac{\theta}{2}}-1=\pm\frac{11}{59}.</cmath>
 +
In this context, <math>\cos{\theta}>0</math>. Thus, <math>\cos{\theta}=\frac{11}{59}</math>. The perimeter of <math>A_1B_1C_1D_1</math> is
 +
<cmath>22\cdot\cos{\theta}=22\cdot\frac{11}{59}=\frac{242}{59},</cmath> and the answer is <math>m+n=242+59=\boxed{301}</math>.
 +
 
 +
~y.grace.yu
 +
 
 +
==Solution 3 (Pythagorean Theorem)==
 +
We assume that the two quadrilateral mentioned in the problem are similar (due to both of them being cyclic). Note that by Ptolemy’s, one of the diagonals has length <math>\sqrt{4 \cdot 6 + 5 \cdot 7} = \sqrt{59}.</math> [I don't believe this is correct... are the two diagonals of <math>ABCD</math> necessarily congruent? -peace09] WLOG we focus on diagonal <math>BD.</math> To find the diagonal of the inner quadrilateral, we drop the altitude from <math>A</math> and <math>C</math> and calculate the length of <math>A_1C_1.</math> Let <math>x</math> be <math>A_1D</math> (Thus <math>A_1B = \sqrt{59} - x.</math> By Pythagorean theorem, we have <cmath>49 - x^2 = 16 - \left(\sqrt{59} - x\right)^2 \implies 92 = 2\sqrt{59}x \implies x = \frac{46}{\sqrt{59}} = \frac{46\sqrt{59}}{59}.</cmath> Now let <math>y</math> be <math>C_1D.</math> (thus making <math>C_1B = \sqrt{59} - y</math>). Similarly, we have <cmath>36 - y^2 = 25 - \left(\sqrt{59} - y\right)^2 \implies 70 = 2\sqrt{59}y \implies y = \frac{35}{\sqrt{59}} = \frac{35\sqrt{59}}{59}.</cmath> We see that <math>A_1C_1</math>, the scaled down diagonal is just <math>x - y = \frac{11\sqrt{59}}{59},</math> which is <math>\frac{\frac{11\sqrt{59}}{59}}{\sqrt{59}} = \frac{11}{59}</math> times our original diagonal <math>BD,</math> implying a scale factor of <math>\frac{11}{59}.</math> Thus, due to perimeters scaling linearly, the perimeter of the new quadrilateral is simply <math>\frac{11}{59} \cdot 22 = \frac{242}{59},</math> making our answer <math>242+59 = \boxed{301}.</math>
 +
 
 +
~fidgetboss_4000
 +
 
 +
==Solution 4 (Symmetry)==
 +
<i><b>Lemma 1</b></i>
 +
[[File:2021 AIME I 11.jpg|500px|right]]
 +
In the triangle <math>ABC</math>, the points <math>B'</math> and <math>C'</math> are the bases of the heights dropped from the vertices <math>B</math> and <math>C</math>, respectively.  <math>\angle A = \alpha</math>. Then <math>B'C' = BC \cos\alpha,</math> <math>\alpha < \frac{\pi}{2}</math> and <math>B'C' = BC \cos (\pi – \alpha), \alpha >\frac{\pi}{2}.</math>
 +
 
 +
<i><b>Proof</b></i>
 +
 
 +
Denote the orthocenter by  <math>A'</math>. Quadrilateral <math>B'C'BC</math> is inscribed in a circle with diameter <math>BC</math>, so the marked <math>\angle B = \angle B'.</math>
 +
 
 +
If  <math>\alpha < \frac {\pi}{2},</math> the <math>\triangle AB'C' \sim \triangle ABC,</math> the similarity coefficient is <math>AC' : AC = \cos \alpha.</math>
 +
So <math>B'C' : BC = \cos \alpha.</math>
 +
 
 +
If  <math>\alpha > \frac {\pi}{2},</math> the <math>\triangle A'B'C' \sim \triangle A'BC,</math> the similarity coefficient is
 +
<math>A'C' : A'C = \cos (\pi – \alpha).</math> So <math>B'C' : BC = \cos (\pi – \alpha).</math>
 +
 
 +
 
 +
 
 +
<i><b>Lemma 2</b></i>
 +
[[File:2021 AIME I 11c.jpg|400px|right]]
 +
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 < \frac{\pi}{2}</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>
 +
<math>\hspace{27mm} d^2 + b^2 – 2 bd \cos \theta = c^2 + a^2 + 2ac \cos \theta,</math>
 +
 
 +
<math>\hspace{29mm} 2(ac + bd) \cos \theta = |d^2 – c^2 + b^2 – a^2|.  </math>
 +
 
 +
 
 +
<i><b>Solution</b></i>
 +
[[File:2021 AIME I 11b.jpg|400px|right]]
 +
In accordance with <b>Lemma 1,</b> the ratios of pairs of one-color segments are the same and equal to <math>\cos \theta,</math> where <math>\theta</math> is the acute angle between the diagonals.
 +
<cmath>s = A'B' + B'C' + C'D' + D'A'=</cmath>
 +
<cmath> = (AB + BC + CD + DA)\cos \theta =</cmath>
 +
<cmath>= (a + b + c + d)\cos \theta = 22\cos \theta.</cmath>
 +
 
 +
In accordance with <b>Lemma 2,</b>
 +
<cmath>\begin{align*} 2(ac + bd)\cos \theta = |d^2 – c^2 + b^2 – a^2|.\end{align*}</cmath>
 +
<cmath>2 \cdot 59 \cos \theta = |13 + 9|.</cmath>
 +
<cmath>s = 22\cos \theta = \frac{22 \cdot 11}{59} = \frac{242}{59}.</cmath>
 +
Therefore, the answer is <math>242+59=\boxed{301}.</math>
 +
 
 +
'''Shelomovskii, vvsss,  www.deoma-cmd.ru'''
 +
 
 +
==See Also==
 
{{AIME box|year=2021|n=I|num-b=10|num-a=12}}
 
{{AIME box|year=2021|n=I|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:10, 21 June 2022

Problem

Let $ABCD$ be a cyclic quadrilateral with $AB=4,BC=5,CD=6,$ and $DA=7.$ Let $A_1$ and $C_1$ be the feet of the perpendiculars from $A$ and $C,$ respectively, to line $BD,$ and let $B_1$ and $D_1$ be the feet of the perpendiculars from $B$ and $D,$ respectively, to line $AC.$ The perimeter of $A_1B_1C_1D_1$ is $\frac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Diagram

[asy] /* Made by MRENTHUSIASM */ size(300);  pair A, B, C, D, A1, B1, C1, D1; A = origin; C = (sqrt(53041)/31,0); B = intersectionpoints(Circle(A,4),Circle(C,5))[0]; D = intersectionpoints(Circle(A,7),Circle(C,6))[1]; A1 = foot(A,B,D); C1 = foot(C,B,D); B1 = foot(B,A,C); D1 = foot(D,A,C); markscalefactor=0.025; draw(rightanglemark(A,A1,B),red); draw(rightanglemark(B,B1,A),red); draw(rightanglemark(C,C1,D),red); draw(rightanglemark(D,D1,C),red); draw(A1--B1--C1--D1--cycle,green); dot("$A$",A,1.5*W,linewidth(4)); dot("$B$",B,1.5*dir(180-aCos(11/59)),linewidth(4)); dot("$C$",C,1.5*E,linewidth(4)); dot("$D$",D,1.5*dir(-aCos(11/59)),linewidth(4)); dot("$A_1$",A1,1.5*dir(A1-A),linewidth(4)); dot("$B_1$",B1,1.5*S,linewidth(4)); dot("$C_1$",C1,1.5*dir(C1-C),linewidth(4)); dot("$D_1$",D1,1.5*N,linewidth(4)); draw(A--B--C--D--cycle^^A--C^^B--D^^circumcircle(A,B,C)); draw(A--A1^^B--B1^^C--C1^^D--D1,dashed); [/asy] ~MRENTHUSIASM

Solution 1 (Cyclic Quadrilaterals, Similar Triangles, and Trigonometry)

This solution refers to the Diagram section.

By the Converse of the Inscribed Angle Theorem, if distinct points $X$ and $Y$ lie on the same side of $\overline{PQ}$ (but not on $\overline{PQ}$ itself) for which $\angle PXQ=\angle PYQ,$ then $P,Q,X,$ and $Y$ are cyclic. From the Converse of the Inscribed Angle Theorem, quadrilaterals $ABA_1B_1,BCC_1B_1,CDC_1D_1,$ and $DAA_1D_1$ are all cyclic.

Suppose $\overline{AC}$ and $\overline{BD}$ intersect at $E,$ and let $\angle AEB=\theta.$ It follows that $\angle CED=\theta$ and $\angle BEC=\angle DEA=180^\circ-\theta.$

We obtain the following diagram: [asy] /* Made by MRENTHUSIASM */ size(300);  pair A, B, C, D, A1, B1, C1, D1, P, M1, M2; A = origin; C = (sqrt(53041)/31,0); B = intersectionpoints(Circle(A,4),Circle(C,5))[0]; D = intersectionpoints(Circle(A,7),Circle(C,6))[1]; A1 = foot(A,B,D); C1 = foot(C,B,D); B1 = foot(B,A,C); D1 = foot(D,A,C); P = intersectionpoint(A--C,B--D); M1 = midpoint(A--B); M2 = midpoint(C--D); markscalefactor=0.025; draw(rightanglemark(A,A1,B),red); draw(rightanglemark(B,B1,A),red); draw(rightanglemark(C,C1,D),red); draw(rightanglemark(D,D1,C),red); draw(Arc(M1,A,B)^^Arc(M2,C,D),blue); draw(A1--B1--C1--D1--cycle,green); dot("$A$",A,1.5*W,linewidth(4)); dot("$B$",B,1.5*dir(180-aCos(11/59)),linewidth(4)); dot("$C$",C,1.5*E,linewidth(4)); dot("$D$",D,1.5*dir(-aCos(11/59)),linewidth(4)); dot("$A_1$",A1,1.5*dir(A1-A),linewidth(4)); dot("$B_1$",B1,1.5*S,linewidth(4)); dot("$C_1$",C1,1.5*dir(C1-C),linewidth(4)); dot("$D_1$",D1,1.5*N,linewidth(4)); dot("$E$",P,dir((180-aCos(11/59))/2),linewidth(4)); label("$\theta$",P,dir(180-aCos(11/59)/2),red); draw(A--B--C--D--cycle^^A--C^^B--D^^circumcircle(A,B,C)); draw(A--A1^^B--B1^^C--C1^^D--D1,dashed); [/asy] In every cyclic quadrilateral, each pair of opposite angles is supplementary. So, we have $\angle EA_1B_1=\angle EAB$ (both supplementary to $\angle B_1A_1B$) and $\angle EB_1A_1=\angle EBA$ (both supplementary to $\angle A_1B_1A$), from which $\triangle A_1B_1E \sim \triangle ABE$ by AA, with the ratio of similitude \[\frac{A_1B_1}{AB}=\underbrace{\frac{A_1E}{AE}}_{\substack{\text{right} \\ \triangle A_1AE}}=\underbrace{\frac{B_1E}{BE}}_{\substack{\text{right} \\ \triangle B_1BE}}=\cos\theta. \hspace{15mm}(1)\] Similarly, we have $\angle EC_1D_1=\angle ECD$ (both supplementary to $\angle D_1C_1D$) and $\angle ED_1C_1=\angle EDC$ (both supplementary to $\angle C_1D_1C$), from which $\triangle C_1D_1E \sim \triangle CDE$ by AA, with the ratio of similitude \[\frac{C_1D_1}{CD}=\underbrace{\frac{C_1E}{CE}}_{\substack{\text{right} \\ \triangle C_1CE}}=\underbrace{\frac{D_1E}{DE}}_{\substack{\text{right} \\ \triangle D_1DE}}=\cos\theta. \hspace{14.75mm}(2)\] We apply the Transitive Property to $(1)$ and $(2):$

  1. We get $\frac{B_1E}{BE}=\frac{C_1E}{CE}=\cos\theta,$ so $\triangle B_1C_1E \sim \triangle BCE$ by SAS, with the ratio of similitude \[\frac{B_1C_1}{BC}=\frac{B_1E}{BE}=\frac{C_1E}{CE}=\cos\theta. \hspace{14.75mm}(3)\]
  2. We get $\frac{D_1E}{DE}=\frac{A_1E}{AE}=\cos\theta,$ so $\triangle D_1A_1E \sim \triangle DAE$ by SAS, with the ratio of similitude \[\frac{D_1A_1}{DA}=\frac{D_1E}{DE}=\frac{A_1E}{AE}=\cos\theta. \hspace{14mm}(4)\]

From $(1),(2),(3),$ and $(4),$ the perimeter of $A_1B_1C_1D_1$ is \begin{align*} A_1B_1+B_1C_1+C_1D_1+D_1A_1&=AB\cos\theta+BC\cos\theta+CD\cos\theta+DA\cos\theta \\ &=(AB+BC+CD+DA)\cos\theta \\ &=22\cos\theta. &&\hspace{5mm}(\bigstar) \end{align*} Two solutions follow from here:

Solution 1.1 (Law of Cosines)

Note that $\cos(180^\circ-\theta)=-\cos\theta$ holds for all $\theta.$ We apply the Law of Cosines to $\triangle ABE, \triangle BCE, \triangle CDE,$ and $\triangle DAE,$ respectively: \begin{alignat*}{12} &&&AE^2+BE^2-2\cdot AE\cdot BE\cdot\cos\angle AEB&&=AB^2&&\quad\implies\quad AE^2+BE^2-2\cdot AE\cdot BE\cdot\cos\theta&&=4^2, \hspace{15mm} &(1\star) \\ &&&BE^2+CE^2-2\cdot BE\cdot CE\cdot\cos\angle BEC&&=BC^2&&\quad\implies\quad BE^2+CE^2+2\cdot BE\cdot CE\cdot\cos\theta&&=5^2, \hspace{15mm} &(2\star) \\ &&&CE^2+DE^2-2\cdot CE\cdot DE\cdot\cos\angle CED&&=CD^2&&\quad\implies\quad CE^2+DE^2-2\cdot CE\cdot DE\cdot\cos\theta&&=6^2, \hspace{15mm} &(3\star) \\ &&&DE^2+AE^2-2\cdot DE\cdot AE\cdot\cos\angle DEA&&=DA^2&&\quad\implies\quad DE^2+AE^2+2\cdot DE\cdot AE\cdot\cos\theta&&=7^2. \hspace{15mm} &(4\star) \\ \end{alignat*} We subtract $(1\star)+(3\star)$ from $(2\star)+(4\star):$ \begin{align*} 2\cdot AE\cdot BE\cdot\cos\theta+2\cdot BE\cdot CE\cdot\cos\theta+2\cdot CE\cdot DE\cdot\cos\theta+2\cdot DE\cdot AE\cdot\cos\theta&=22 \\ 2\cdot\cos\theta\cdot(\phantom{ }\underbrace{AE\cdot BE+BE\cdot CE+CE\cdot DE+DE\cdot AE}_{\text{Use the result from }\textbf{Remark}\text{.}}\phantom{ })&=22 \\ 2\cdot\cos\theta\cdot59&=22 \\ \cos\theta&=\frac{11}{59}. \end{align*} Finally, substituting this result into $(\bigstar)$ gives $22\cos\theta=\frac{242}{59},$ from which the answer is $242+59=\boxed{301}.$

~MRENTHUSIASM (credit given to Math Jams's 2021 AIME I Discussion)

Solution 1.2 (Area Formulas)

Let the brackets denote areas. We find $[ABCD]$ in two different ways:

  1. Note that $\sin(180^\circ-\theta)=\sin\theta$ holds for all $\theta.$ By area addition, we get \begin{align*} [ABCD]&=[ABE]+[BCE]+[CDE]+[DAE] \\ &=\frac12\cdot AE\cdot BE\cdot\sin\angle AEB+\frac12\cdot BE\cdot CE\cdot\sin\angle BEC+\frac12\cdot CE\cdot DE\cdot\sin\angle CED+\frac12\cdot DE\cdot AE\cdot\sin\angle DEA \\ &=\frac12\cdot AE\cdot BE\cdot\sin\theta+\frac12\cdot BE\cdot CE\cdot\sin\theta+\frac12\cdot CE\cdot DE\cdot\sin\theta+\frac12\cdot DE\cdot AE\cdot\sin\theta \\ &=\frac12\cdot\sin\theta\cdot(\phantom{ }\underbrace{AE\cdot BE+BE\cdot CE+CE\cdot DE+DE\cdot AE}_{\text{Use the result from }\textbf{Remark}\text{.}}\phantom{ }) \\ &=\frac12\cdot\sin\theta\cdot59. \end{align*}
  2. By Brahmagupta's Formula, we get \[[ABCD]=\sqrt{(s-AB)(s-BC)(s-CD)(s-DA)}=2\sqrt{210},\] where $s=\frac{AB+BC+CD+DA}{2}=11$ is the semiperimeter of $ABCD.$

Equating the expressions for $[ABCD],$ we have \[\frac12\cdot\sin\theta\cdot59=2\sqrt{210},\] so $\sin\theta=\frac{4\sqrt{210}}{59}.$ Since $0^\circ<\theta<90^\circ,$ we have $\cos\theta>0.$ It follows that \[\cos\theta=\sqrt{1-\sin^2\theta}=\frac{11}{59}.\] Finally, substituting this result into $(\bigstar)$ gives $22\cos\theta=\frac{242}{59},$ from which the answer is $242+59=\boxed{301}.$

~MRENTHUSIASM (credit given to Leonard my dude)

Remark (Ptolemy's Theorem)

In $ABCD,$ we have \begin{align*} AE\cdot BE+BE\cdot CE+CE\cdot DE+DE\cdot AE &= (AE+CE)(BE+DE) &&\hspace{10mm}\text{Factor by Grouping} \\ &=AC\cdot BD &&\hspace{10mm}\text{Segment Addition} \\ &=AB\cdot CD+BC\cdot DA &&\hspace{10mm}\text{Ptolemy's Theorem} \\ &=59. &&\hspace{10mm}\text{Substitution} \end{align*} ~MRENTHUSIASM

Solution 2 (Finding cos x)

The angle $\theta$ between diagonals satisfies \[\tan{\frac{\theta}{2}}=\sqrt{\frac{(s-b)(s-d)}{(s-a)(s-c)}}\] (see https://en.wikipedia.org/wiki/Cyclic_quadrilateral#Angle_formulas). Thus, \[\tan{\frac{\theta}{2}}=\sqrt{\frac{(11-4)(11-6)}{(11-5)(11-7)}}\text{ or }\tan{\frac{\theta}{2}}=\sqrt{\frac{(11-5)(11-7)}{(11-4)(11-6)}}.\] That is, $\tan^2{\frac{\theta}{2}}=\frac{1-\cos^2{\frac{\theta}{2}}}{\cos^2{\frac{\theta}{2}}}=\frac{24}{35}$ or $\frac{35}{24}$. Thus, $\cos^2{\frac{\theta}{2}}=\frac{35}{59}$ or $\frac{24}{59}$. So, \[\cos{\theta}=2\cos^2{\frac{\theta}{2}}-1=\pm\frac{11}{59}.\] In this context, $\cos{\theta}>0$. Thus, $\cos{\theta}=\frac{11}{59}$. The perimeter of $A_1B_1C_1D_1$ is \[22\cdot\cos{\theta}=22\cdot\frac{11}{59}=\frac{242}{59},\] and the answer is $m+n=242+59=\boxed{301}$.

~y.grace.yu

Solution 3 (Pythagorean Theorem)

We assume that the two quadrilateral mentioned in the problem are similar (due to both of them being cyclic). Note that by Ptolemy’s, one of the diagonals has length $\sqrt{4 \cdot 6 + 5 \cdot 7} = \sqrt{59}.$ [I don't believe this is correct... are the two diagonals of $ABCD$ necessarily congruent? -peace09] WLOG we focus on diagonal $BD.$ To find the diagonal of the inner quadrilateral, we drop the altitude from $A$ and $C$ and calculate the length of $A_1C_1.$ Let $x$ be $A_1D$ (Thus $A_1B = \sqrt{59} - x.$ By Pythagorean theorem, we have \[49 - x^2 = 16 - \left(\sqrt{59} - x\right)^2 \implies 92 = 2\sqrt{59}x \implies x = \frac{46}{\sqrt{59}} = \frac{46\sqrt{59}}{59}.\] Now let $y$ be $C_1D.$ (thus making $C_1B = \sqrt{59} - y$). Similarly, we have \[36 - y^2 = 25 - \left(\sqrt{59} - y\right)^2 \implies 70 = 2\sqrt{59}y \implies y = \frac{35}{\sqrt{59}} = \frac{35\sqrt{59}}{59}.\] We see that $A_1C_1$, the scaled down diagonal is just $x - y = \frac{11\sqrt{59}}{59},$ which is $\frac{\frac{11\sqrt{59}}{59}}{\sqrt{59}} = \frac{11}{59}$ times our original diagonal $BD,$ implying a scale factor of $\frac{11}{59}.$ Thus, due to perimeters scaling linearly, the perimeter of the new quadrilateral is simply $\frac{11}{59} \cdot 22 = \frac{242}{59},$ making our answer $242+59 = \boxed{301}.$

~fidgetboss_4000

Solution 4 (Symmetry)

Lemma 1

2021 AIME I 11.jpg

In the triangle $ABC$, the points $B'$ and $C'$ are the bases of the heights dropped from the vertices $B$ and $C$, respectively. $\angle A = \alpha$. Then $B'C' = BC \cos\alpha,$ $\alpha < \frac{\pi}{2}$ and $B'C' = BC \cos (\pi – \alpha), \alpha >\frac{\pi}{2}.$

Proof

Denote the orthocenter by $A'$. Quadrilateral $B'C'BC$ is inscribed in a circle with diameter $BC$, so the marked $\angle B = \angle B'.$

If $\alpha < \frac {\pi}{2},$ the $\triangle AB'C' \sim \triangle ABC,$ the similarity coefficient is $AC' : AC = \cos \alpha.$ So $B'C' : BC = \cos \alpha.$

If $\alpha > \frac {\pi}{2},$ the $\triangle A'B'C' \sim \triangle A'BC,$ the similarity coefficient is $A'C' : A'C = \cos (\pi – \alpha).$ So $B'C' : BC = \cos (\pi – \alpha).$


Lemma 2

2021 AIME I 11c.jpg

Given an inscribed quadrilateral $ABCD$ with sides $AB = a, BC = b, CD = c,$ and $DA = d.$ Prove that the $\angle \theta < \frac{\pi}{2}$ between the diagonals is given by \begin{align*}2(ac + bd) \cos \theta =  {|d^2 – c^2 + b^2 – a^2|}.\end{align*} Proof

Let the point $B'$ be symmetric to $B$ with respect to the perpendicular bisector $AC.$ Then the quadrilateral $AB'CD$ is an inscribed one, $AB' = b, B'C = a.$

\[2 \angle AEB = \overset{\Large\frown} {AB}  + \overset{\Large\frown} {CD}.\] \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*}

We apply the Law of Cosines to $\triangle AB'D$ and $\triangle CB'D$: \begin{align*} B'D^2 = AD^2 + AB'^2 – 2 AD \cdot AB' \cos \theta,\end{align*} \begin{align*} B'D^2 = CD^2 + CB'^2 + 2 CD \cdot CB' \cos \theta,\end{align*} $\hspace{27mm} d^2 + b^2 – 2 bd \cos \theta = c^2 + a^2 + 2ac \cos \theta,$

$\hspace{29mm} 2(ac + bd) \cos \theta = |d^2 – c^2 + b^2 – a^2|.$


Solution

2021 AIME I 11b.jpg

In accordance with Lemma 1, the ratios of pairs of one-color segments are the same and equal to $\cos \theta,$ where $\theta$ is the acute angle between the diagonals. \[s = A'B' + B'C' + C'D' + D'A'=\] \[= (AB + BC + CD + DA)\cos \theta =\] \[= (a + b + c + d)\cos \theta = 22\cos \theta.\]

In accordance with Lemma 2, \begin{align*} 2(ac + bd)\cos \theta = |d^2 – c^2 + b^2 – a^2|.\end{align*} \[2 \cdot 59 \cos \theta = |13 + 9|.\] \[s = 22\cos \theta = \frac{22 \cdot 11}{59} = \frac{242}{59}.\] Therefore, the answer is $242+59=\boxed{301}.$

Shelomovskii, vvsss, www.deoma-cmd.ru

See Also

2021 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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All AIME Problems and Solutions

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