Difference between revisions of "2021 AIME I Problems/Problem 11"
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==Problem== | ==Problem== | ||
− | Let <math>ABCD</math> be a cyclic quadrilateral with <math>AB=4,BC=5,CD=6,</math> and <math>DA=7</math> | + | Let <math>ABCD</math> be a cyclic quadrilateral with <math>AB=4,BC=5,CD=6,</math> and <math>DA=7.</math> Let <math>A_1</math> and <math>C_1</math> be the feet of the perpendiculars from <math>A</math> and <math>C,</math> respectively, to line <math>BD,</math> and let <math>B_1</math> and <math>D_1</math> be the feet of the perpendiculars from <math>B</math> and <math>D,</math> respectively, to line <math>AC.</math> The perimeter of <math>A_1B_1C_1D_1</math> is <math>\frac mn,</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n.</math> |
==Diagram== | ==Diagram== | ||
− | + | <asy> | |
+ | /* Made by MRENTHUSIASM */ | ||
+ | size(300); | ||
− | ~MRENTHUSIASM | + | 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 | + | ==Solution 1 (Cyclic Quadrilaterals, Similar Triangles, and Trigonometry)== |
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This solution refers to the <b>Diagram</b> section. | This solution refers to the <b>Diagram</b> section. | ||
− | + | By the <b>Converse of the Inscribed Angle Theorem</b>, if distinct points <math>X</math> and <math>Y</math> lie on the same side of <math>\overline{PQ}</math> (but not on <math>\overline{PQ}</math> itself) for which <math>\angle PXQ=\angle PYQ,</math> then <math>P,Q,X,</math> and <math>Y</math> are cyclic. From the Converse of the Inscribed Angle Theorem, quadrilaterals <math>ABA_1B_1,BCC_1B_1,CDC_1D_1,</math> and <math>DAA_1D_1</math> are all cyclic. | |
− | + | Suppose <math>\overline{AC}</math> and <math>\overline{BD}</math> intersect at <math>E,</math> and let <math>\angle AEB=\theta.</math> It follows that <math>\angle CED=\theta</math> and <math>\angle BEC=\angle DEA=180^\circ-\theta.</math> | |
We obtain the following diagram: | We obtain the following diagram: | ||
− | [[ | + | <asy> |
− | In every cyclic quadrilateral, each pair of opposite angles is supplementary. So, we have <math>\angle EA_1B_1=\angle EAB</math> and <math>\angle EB_1A_1=\angle EBA</math> | + | /* Made by MRENTHUSIASM */ |
− | Similarly, we have <math>\angle EC_1D_1=\angle ECD</math> and <math>\angle ED_1C_1=\angle EDC</math> | + | 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> | ||
+ | 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> | ||
We apply the Transitive Property to <math>(1)</math> and <math>(2):</math> | We apply the Transitive Property to <math>(1)</math> and <math>(2):</math> | ||
<ol style="margin-left: 1.5em;"> | <ol style="margin-left: 1.5em;"> | ||
− | <li>We get <math>\frac{B_1E}{BE}=\frac{C_1E}{CE}=\cos\theta,</math> | + | <li>We get <math>\frac{B_1E}{BE}=\frac{C_1E}{CE}=\cos\theta,</math> so <math>\triangle B_1C_1E \sim \triangle BCE</math> by SAS, with the ratio of similitude <cmath>\frac{B_1C_1}{BC}=\frac{B_1E}{BE}=\frac{C_1E}{CE}=\cos\theta. \hspace{14.75mm}(3)</cmath></li><p> |
− | <li>We get <math>\frac{D_1E}{DE}=\frac{A_1E}{AE}=\cos\theta,</math> | + | <li>We get <math>\frac{D_1E}{DE}=\frac{A_1E}{AE}=\cos\theta,</math> so <math>\triangle D_1A_1E \sim \triangle DAE</math> by SAS, with the ratio of similitude <cmath>\frac{D_1A_1}{DA}=\frac{D_1E}{DE}=\frac{A_1E}{AE}=\cos\theta. \hspace{14mm}(4)</cmath></li><p> |
</ol> | </ol> | ||
− | From <math>(1),(2),(3),</math> and <math>(4),</math> | + | From <math>(1),(2),(3),</math> and <math>(4),</math> the perimeter of <math>A_1B_1C_1D_1</math> is |
<cmath>\begin{align*} | <cmath>\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 \\ | 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 \\ | &=(AB+BC+CD+DA)\cos\theta \\ | ||
− | &=22\cos\theta. \hspace{ | + | &=22\cos\theta. &&\hspace{5mm}(\bigstar) |
\end{align*}</cmath> | \end{align*}</cmath> | ||
+ | Two solutions follow from here: | ||
+ | |||
+ | ===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*}{ | + | <cmath>\begin{alignat*}{12} |
− | AB^2& | + | &&&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) \\ |
− | BC^2& | + | &&&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) \\ |
− | CD^2& | + | &&&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) \\ |
− | DA^2& | + | &&&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*}</cmath> | \end{alignat*}</cmath> | ||
− | We subtract <math>(1\star)+(3\star)</math> from <math>(2\star)+(4\star) | + | We subtract <math>(1\star)+(3\star)</math> from <math>(2\star)+(4\star):</math> |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | 2\cdot\cos\theta\cdot | + | 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 \\ |
− | \cos\theta\cdot(AE\cdot BE+BE\cdot CE+CE\cdot DE+DE\cdot AE | + | 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\cdot59&= | ||
\cos\theta&=\frac{11}{59}. | \cos\theta&=\frac{11}{59}. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
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 Math Jams's <b>2021 AIME I Discussion</b>) |
+ | |||
+ | ===Solution 1.2 (Area Formulas)=== | ||
+ | Let the brackets denote areas. | ||
+ | We find <math>[ABCD]</math> in two different ways: | ||
+ | <ol style="margin-left: 1.5em;"> | ||
+ | <li>Note that <math>\sin(180^\circ-\theta)=\sin\theta</math> holds for all <math>\theta.</math> By area addition, we get | ||
+ | <cmath>\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*}</cmath></li><p> | ||
+ | <li>By Brahmagupta's Formula, we get <cmath>[ABCD]=\sqrt{(s-AB)(s-BC)(s-CD)(s-DA)}=2\sqrt{210},</cmath> where <math>s=\frac{AB+BC+CD+DA}{2}=11</math> is the semiperimeter of <math>ABCD.</math></li><p> | ||
+ | </ol> | ||
+ | Equating the expressions for <math>[ABCD],</math> we have | ||
+ | <cmath>\frac12\cdot\sin\theta\cdot59=2\sqrt{210},</cmath> so <math>\sin\theta=\frac{4\sqrt{210}}{59}.</math> Since <math>0^\circ<\theta<90^\circ,</math> we have <math>\cos\theta>0.</math> It follows that <cmath>\cos\theta=\sqrt{1-\sin^2\theta}=\frac{11}{59}.</cmath> | ||
+ | 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 (credit given to Leonard my dude) | ||
+ | |||
+ | ===Remark (Ptolemy's Theorem)=== | ||
+ | In <math>ABCD,</math> we have | ||
+ | <cmath>\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*}</cmath> | ||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | ==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 | ||
+ | |||
+ | <cmath></cmath> | ||
+ | <math>*</math>Indeed, the diagonals do not have length <math>\sqrt{59}</math>. However, here's why it works out: Let the length of diagonal BD be a. Then, by the same logic as given in the solution by fidgetboss_4000, <math>x=\frac{33+a^{2}}{2a}</math>, <math>y=\frac{11+a^{2}}{2a}</math>, so <math>x-y=\frac{11}{a}</math>. Then, why does dividing by a to get a ratio work? It's because the orientation of quadrilaterals is different! In other words, instead of diagonal BD corresponding to diagonal <math>B_1</math> <math>D_1</math>, it corresponds to diagonal <math>A_1</math> <math>C_1</math>. Thus, to get the right ratio, we can find the ratio of the multiplied diagonals, then take the square root (this square root part is crucial. It happens because we essentially are taking something two-dimensional on the numerator and denominator, so to make it a one-dimensional similarity ratio, we must take the square root to half the dimension on each part). This yields <math>\sqrt{\frac{121}{a^{2}\cdot b^{2}}}</math>=<math>\frac{11}{a\cdot b}</math>. However, by Ptolemy, <math>a\cdot b</math>=59. Thus, the ratio is indeed <math>\frac{11}{59}</math>. ~MATH-TITAN | ||
+ | |||
+ | ==Solution 4 (Symmetry)== | ||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | [[File:AIME-I-2021-11a.png|350px|right]] | ||
+ | |||
+ | In accordance with <b>Claim 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>\begin{align*} | ||
+ | s &= A'B' + B'C' + C'D' + D'A' \\ | ||
+ | &= (AB + BC + CD + DA)\cos \theta \\ | ||
+ | &= (a + b + c + d)\cos \theta \\ | ||
+ | &= 22\cos \theta. | ||
+ | \end{align*}</cmath> | ||
+ | In accordance with <b>Claim 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> | ||
+ | |||
+ | <i><b>Claim 1</b></i> | ||
+ | [[File:AIME-I-2021-11b.png|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 | ||
+ | <cmath>B'C'=\begin{cases} | ||
+ | BC \cos\alpha, & \textrm{if } \alpha < 90^\circ, \\ | ||
+ | BC \cos (180^\circ – \alpha), & \textrm{if } \alpha >90^\circ. | ||
+ | \end{cases}=|BC\cos{\alpha}| | ||
+ | </cmath> | ||
+ | |||
+ | <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 < 90^\circ,</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 > 90^\circ,</math> the <math>\triangle A'B'C' \sim \triangle A'BC,</math> the similarity coefficient is | ||
+ | <math>A'C' : A'C = \cos (180^\circ – \alpha).</math> So <math>B'C' : BC = \cos (180^\circ – \alpha).</math> | ||
+ | |||
+ | <i><b>Claim 2</b></i> | ||
+ | [[File:AIME-I-2021-11c.png|300px|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 < 90^\circ</math> between the diagonals is given by | ||
+ | <cmath>\begin{align*} \cos \theta = \frac {|d^2 – c^2 + b^2 – a^2|}{2(ac + bd)}.\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>B'D^2 = CD^2 + CB'^2 + 2 CD \cdot CB' \cos \theta,</cmath> | ||
+ | <cmath>d^2 + b^2 – 2 bd \cos \theta = c^2 + a^2 + 2ac \cos \theta,</cmath> | ||
+ | <cmath>2(ac + bd) \cos \theta = |d^2 – c^2 + b^2 – a^2|.</cmath> | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Note== | ||
+ | This problem is kinda similar to [[2021 AIME II Problems/Problem 12]] | ||
− | ==See | + | ==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}} |
Latest revision as of 19:56, 24 December 2023
Contents
Problem
Let be a cyclic quadrilateral with
and
Let
and
be the feet of the perpendiculars from
and
respectively, to line
and let
and
be the feet of the perpendiculars from
and
respectively, to line
The perimeter of
is
where
and
are relatively prime positive integers. Find
Diagram
~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 and
lie on the same side of
(but not on
itself) for which
then
and
are cyclic. From the Converse of the Inscribed Angle Theorem, quadrilaterals
and
are all cyclic.
Suppose and
intersect at
and let
It follows that
and
We obtain the following diagram:
In every cyclic quadrilateral, each pair of opposite angles is supplementary. So, we have
(both supplementary to
) and
(both supplementary to
), from which
by AA, with the ratio of similitude
Similarly, we have
(both supplementary to
) and
(both supplementary to
), from which
by AA, with the ratio of similitude
We apply the Transitive Property to
and
- We get
so
by SAS, with the ratio of similitude
- We get
so
by SAS, with the ratio of similitude
From and
the perimeter of
is
Two solutions follow from here:
Solution 1.1 (Law of Cosines)
Note that holds for all
We apply the Law of Cosines to
and
respectively:
We subtract
from
Finally, substituting this result into
gives
from which the answer is
~MRENTHUSIASM (credit given to Math Jams's 2021 AIME I Discussion)
Solution 1.2 (Area Formulas)
Let the brackets denote areas.
We find in two different ways:
- Note that
holds for all
By area addition, we get
- By Brahmagupta's Formula, we get
where
is the semiperimeter of
Equating the expressions for we have
so
Since
we have
It follows that
Finally, substituting this result into
gives
from which the answer is
~MRENTHUSIASM (credit given to Leonard my dude)
Remark (Ptolemy's Theorem)
In we have
~MRENTHUSIASM
Solution 2 (Finding cos x)
The angle between diagonals satisfies
(see https://en.wikipedia.org/wiki/Cyclic_quadrilateral#Angle_formulas).
Thus,
That is,
or
. Thus,
or
. So,
In this context,
. Thus,
. The perimeter of
is
and the answer is
.
~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 [I don't believe this is correct... are the two diagonals of
necessarily congruent? -peace09]* WLOG we focus on diagonal
To find the diagonal of the inner quadrilateral, we drop the altitude from
and
and calculate the length of
Let
be
(Thus
By Pythagorean theorem, we have
Now let
be
(thus making
). Similarly, we have
We see that
, the scaled down diagonal is just
which is
times our original diagonal
implying a scale factor of
Thus, due to perimeters scaling linearly, the perimeter of the new quadrilateral is simply
making our answer
~fidgetboss_4000
Indeed, the diagonals do not have length
. However, here's why it works out: Let the length of diagonal BD be a. Then, by the same logic as given in the solution by fidgetboss_4000,
,
, so
. Then, why does dividing by a to get a ratio work? It's because the orientation of quadrilaterals is different! In other words, instead of diagonal BD corresponding to diagonal
, it corresponds to diagonal
. Thus, to get the right ratio, we can find the ratio of the multiplied diagonals, then take the square root (this square root part is crucial. It happens because we essentially are taking something two-dimensional on the numerator and denominator, so to make it a one-dimensional similarity ratio, we must take the square root to half the dimension on each part). This yields
=
. However, by Ptolemy,
=59. Thus, the ratio is indeed
. ~MATH-TITAN
Solution 4 (Symmetry)
Solution
In accordance with Claim 1, the ratios of pairs of one-color segments are the same and equal to where
is the acute angle between the diagonals.
In accordance with Claim 2,
Therefore, the answer is
Claim 1
In the triangle , the points
and
are the bases of the heights dropped from the vertices
and
, respectively.
. Then
Proof
Denote the orthocenter by . Quadrilateral
is inscribed in a circle with diameter
, so the marked
If the
the similarity coefficient is
So
If the
the similarity coefficient is
So
Claim 2
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
This problem is kinda similar to 2021 AIME II Problems/Problem 12
See Also
2021 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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