Difference between revisions of "2019 AMC 12B Problems/Problem 25"
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==Problem== | ==Problem== | ||
− | Let <math>ABCD</math> be a convex quadrilateral with <math>BC=2</math> and <math>CD=6.</math> Suppose that the centroids of <math>\triangle ABC,\triangle BCD,</math> and <math>\triangle ACD</math> form the vertices of an equilateral triangle. What is the maximum possible value of <math>ABCD</math>? | + | Let <math>ABCD</math> be a convex quadrilateral with <math>BC=2</math> and <math>CD=6.</math> Suppose that the centroids of <math>\triangle ABC,\triangle BCD,</math> and <math>\triangle ACD</math> form the vertices of an equilateral triangle. What is the maximum possible value of the area of <math>ABCD</math>? |
<math>\textbf{(A) } 27 \qquad\textbf{(B) } 16\sqrt3 \qquad\textbf{(C) } 12+10\sqrt3 \qquad\textbf{(D) } 9+12\sqrt3 \qquad\textbf{(E) } 30</math> | <math>\textbf{(A) } 27 \qquad\textbf{(B) } 16\sqrt3 \qquad\textbf{(C) } 12+10\sqrt3 \qquad\textbf{(D) } 9+12\sqrt3 \qquad\textbf{(E) } 30</math> | ||
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Hence <math>\overrightarrow{G_{1}G_{2}} = \frac{1}{3}\vec{q}</math>, <math>\overrightarrow{G_{2}G_{3}} = -\frac{1}{3}\vec{p}</math>, and <math>\overrightarrow{G_{3}G_{1}} = \frac{1}{3}\vec{p} - \frac{1}{3}\vec{q}</math>. For <math>\triangle G_{1}G_{2}G_{3}</math> to be equilateral, we need <math>\left|\overrightarrow{G_{1}G_{2}}\right| = \left|\overrightarrow{G_{2}G_{3}}\right| \Rightarrow \left|\vec{p}\right| = \left|\vec{q}\right| \Rightarrow AB = AD</math>. Further, <math>\left|\overrightarrow{G_{1}G_{2}}\right| = \left|\overrightarrow{G_{1}G_{3}}\right| \Rightarrow \left|\vec{p}\right| = \left|\vec{p} - \vec{q}\right| = BD</math>. Hence we have <math>AB = AD = BD</math>, so <math>\triangle ABD</math> is equilateral. | Hence <math>\overrightarrow{G_{1}G_{2}} = \frac{1}{3}\vec{q}</math>, <math>\overrightarrow{G_{2}G_{3}} = -\frac{1}{3}\vec{p}</math>, and <math>\overrightarrow{G_{3}G_{1}} = \frac{1}{3}\vec{p} - \frac{1}{3}\vec{q}</math>. For <math>\triangle G_{1}G_{2}G_{3}</math> to be equilateral, we need <math>\left|\overrightarrow{G_{1}G_{2}}\right| = \left|\overrightarrow{G_{2}G_{3}}\right| \Rightarrow \left|\vec{p}\right| = \left|\vec{q}\right| \Rightarrow AB = AD</math>. Further, <math>\left|\overrightarrow{G_{1}G_{2}}\right| = \left|\overrightarrow{G_{1}G_{3}}\right| \Rightarrow \left|\vec{p}\right| = \left|\vec{p} - \vec{q}\right| = BD</math>. Hence we have <math>AB = AD = BD</math>, so <math>\triangle ABD</math> is equilateral. | ||
− | Now let the side length of <math>\triangle ABD</math> be <math>k</math>, and let <math>\angle BCD = \theta</math>. By the Law of Cosines in <math>\triangle BCD</math>, we have <math>k^2 = 2^2 + 6^2 - 2 \cdot 2 \cdot 6 \cdot \cos{\theta} = 40 - 24\cos{\theta}</math>. Since <math>\triangle ABD</math> is equilateral, its area is <math>\frac{\sqrt{3}}{4}k^2 = 10\sqrt{3} - 6\sqrt{3}\cos{\theta}</math>, while the area of <math>\triangle BCD</math> is <math>\frac{1}{2} \cdot 2 \cdot 6 \cdot \sin{\theta} = 6 \sin{\theta}</math>. Thus the total area of <math>ABCD</math> is <math>10\sqrt{3} + 6\left(\sin{\theta} - \sqrt{3}\cos{\theta}\right) = 10\sqrt{3} + 12\left( | + | Now let the side length of <math>\triangle ABD</math> be <math>k</math>, and let <math>\angle BCD = \theta</math>. By the Law of Cosines in <math>\triangle BCD</math>, we have <math>k^2 = 2^2 + 6^2 - 2 \cdot 2 \cdot 6 \cdot \cos{\theta} = 40 - 24\cos{\theta}</math>. Since <math>\triangle ABD</math> is equilateral, its area is <math>\frac{\sqrt{3}}{4}k^2 = 10\sqrt{3} - 6\sqrt{3}\cos{\theta}</math>, while the area of <math>\triangle BCD</math> is <math>\frac{1}{2} \cdot 2 \cdot 6 \cdot \sin{\theta} = 6 \sin{\theta}</math>. Thus the total area of <math>ABCD</math> is <math>10\sqrt{3} + 6\left(\sin{\theta} - \sqrt{3}\cos{\theta}\right) = 10\sqrt{3} + 12\left(\frac{1}{2} \sin{\theta} - \frac{\sqrt{3}}{2}\cos{\theta}\right) = 10\sqrt{3}+12\sin{\left(\theta-60^{\circ}\right)}</math>, where in the last step we used the subtraction formula for <math>\sin</math>. Alternatively, we can use calculus to find the local maximum. Observe that <math>\sin{\left(\theta-60^{\circ}\right)}</math> has maximum value <math>1</math> when e.g. <math>\theta = 150^{\circ}</math>, which is a valid configuration, so the maximum area is <math>10\sqrt{3} + 12(1) = \boxed{\textbf{(C) } 12+10\sqrt3}</math>. |
==Solution 2== | ==Solution 2== | ||
− | Let <math>G_1</math>, <math>G_2</math>, <math>G_3</math> be the centroids of <math>ABC</math>, <math>BCD</math>, and <math>CDA</math> respectively, and let <math>M</math> be the midpoint of <math>BC</math>. <math>A</math>, <math>G_1</math>, and <math>M</math> are collinear due to well-known properties of the centroid. Likewise, <math>D</math>, <math>G_2</math>, and <math>M</math> are collinear as well. Because (as is also well-known) <math> | + | Let <math>G_1</math>, <math>G_2</math>, <math>G_3</math> be the centroids of <math>ABC</math>, <math>BCD</math>, and <math>CDA</math> respectively, and let <math>M</math> be the midpoint of <math>BC</math>. <math>A</math>, <math>G_1</math>, and <math>M</math> are collinear due to well-known properties of the centroid. Likewise, <math>D</math>, <math>G_2</math>, and <math>M</math> are collinear as well. Because (as is also well-known) <math>3MG_1 = AM</math> and <math>3MG_2 = DM</math>, we have <math>\triangle MG_1G_2\sim\triangle MAD</math>. This implies that <math>AD</math> is parallel to <math>G_1G_2</math>, and in terms of lengths, <math>AD = 3G_1G_2</math>. (SAS Similarity) |
We can apply the same argument to the pair of triangles <math>\triangle BCD</math> and <math>\triangle ACD</math>, concluding that <math>AB</math> is parallel to <math>G_2G_3</math> and <math>AB = 3G_2G_3</math>. Because <math>3G_1G_2 = 3G_2G_3</math> (due to the triangle being equilateral), <math>AB = AD</math>, and the pair of parallel lines preserve the <math>60^{\circ}</math> angle, meaning <math>\angle BAD = 60^\circ</math>. Therefore <math>\triangle BAD</math> is equilateral. | We can apply the same argument to the pair of triangles <math>\triangle BCD</math> and <math>\triangle ACD</math>, concluding that <math>AB</math> is parallel to <math>G_2G_3</math> and <math>AB = 3G_2G_3</math>. Because <math>3G_1G_2 = 3G_2G_3</math> (due to the triangle being equilateral), <math>AB = AD</math>, and the pair of parallel lines preserve the <math>60^{\circ}</math> angle, meaning <math>\angle BAD = 60^\circ</math>. Therefore <math>\triangle BAD</math> is equilateral. | ||
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Thus the maximum possible area of <math>ABCD</math> is <math>\boxed{\textbf{(C) }12 + 10\sqrt{3}}</math>. | Thus the maximum possible area of <math>ABCD</math> is <math>\boxed{\textbf{(C) }12 + 10\sqrt{3}}</math>. | ||
+ | |||
+ | ==Solution 3 (Complex Numbers)== | ||
+ | Let <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math> correspond to the complex numbers <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math>, respectively. Then, the complex representations of the centroids are <math>(a+b+c)/3</math>, <math>(b+c+d)/3</math>, and <math>(a+c+d)/3</math>. The pairwise distances between the centroids are <math>\lvert (d-a)/3 \rvert</math>, <math>\lvert (b-a)/3 \rvert</math>, and <math>\lvert (b-d)/3 \rvert</math>, all equal. Thus, <math>\lvert (b-a)/3 \rvert=\lvert (d-a)/3 \rvert=\lvert (b-d)/3 \rvert</math>, so <math>\lvert (b-a) \rvert=\lvert (d-a) \rvert=\lvert (b-d) \rvert</math>. Hence, <math>\triangle DBA</math> is equilateral. | ||
+ | |||
+ | By the Law of Cosines, | ||
+ | <math>[ABCD]=[ABD]+[BCD]=\frac{(\sqrt{2^2+6^2-2 \cdot 2 \cdot 6 \cos(\angle BCD)})^2 \cdot \sqrt{3}}{4}+1/2 \cdot 2 \cdot 6 \sin(\angle BCD)</math>. | ||
+ | |||
+ | <math>[ABCD]=10\sqrt{3}+6(\sin{\angle BCD}-\sqrt{3}\cos(\angle BCD))= 10\sqrt{3}+12\sin(\angle BCD-60^{\circ}) \le 12 + 10\sqrt{3}</math>. Thus, the maximum possible area of <math>ABCD</math> is <math>\boxed{\textbf{(C) }12 + 10\sqrt{3}}</math>. | ||
+ | |||
+ | ~ Leo.Euler | ||
+ | |||
+ | ==Solution 4 (Homothety)== | ||
+ | Let <math>G_1, G_2</math>, and <math>G_3</math> be the centroids of <math>\triangle ABC, \triangle BCD</math>, and <math>\triangle ACD</math>, respectively, and let <math>X, Y,</math> and <math>Z</math> be the midpoints of <math>\overline{AB}, \overline{BD},</math> and <math>\overline{AD}</math>, respectively. Note that <math>G_1, G_2,</math> and <math>G_3</math> are <math>\frac{2}{3}</math> of the way from <math>C</math> to <math>X, Y,</math> and <math>Z</math>, respectively, by a well-known property of centroids. Then a homothety centered at <math>C</math> with ratio <math>\frac{3}{2}</math> maps <math>G_1, G_2,</math> and <math>G_3</math> to <math>X, Y,</math> and <math>Z</math>, respectively, implying that <math>\triangle XYZ</math> is equilateral too. But <math>\triangle XYZ</math> is the medial triangle of <math>\triangle ABD</math>, so <math>\triangle ABD</math> is also equilateral. We may finish with the methods in the solutions above. | ||
+ | |||
+ | ~ numberwhiz | ||
+ | |||
+ | While the solutions above have attempted the problem in general, knowing the fact that <math>\triangle ABD</math> is equilateral greatly reduces the effort to find the final answer, hence I propose an alternative after this. | ||
+ | |||
+ | Let <math>AB = BD = AD = x</math> and <math>\angle BCD = \theta</math>. By cosine rule on <math>\triangle BCD</math> : | ||
+ | <cmath>x^2 = 40 - 24\cos \theta</cmath> | ||
+ | Thus, the total area of the quadrilateral is supposedly : | ||
+ | <cmath>\frac{\sqrt{3}}{4}(x^2) + \frac{1}{2}(2)(6)\sin \theta</cmath> | ||
+ | <cmath>\implies \frac{\sqrt{3}}{4}(40 - 24\cos \theta) + 6\sin \theta</cmath> | ||
+ | <cmath>\implies 6(\sin \theta - \sqrt{3}\cos \theta) + 10\sqrt{3} \geq 12 + 10\sqrt{3}</cmath> | ||
+ | Where the inequality comes from a common trigonometric identity, <math>(\sin \theta - \sqrt{3}\cos \theta) \geq \sqrt{1^2 + \big(\sqrt{3}\big)^2} = 2.</math> | ||
+ | |||
+ | ~ SouradipClash_03 | ||
+ | |||
+ | ==Video Solution by MOP 2024== | ||
+ | https://youtu.be/c26N2w2MMQE | ||
+ | |||
+ | ~r00tsOfUnity | ||
+ | |||
+ | ===Solution 5=== | ||
+ | |||
+ | |||
+ | Let <math>X, Y, Z</math> be the centroids of <math>\triangle ABC, \triangle BCD, \triangle ACD</math> respectively, then | ||
+ | |||
+ | <math>XZ//BD</math>, since <math>EX=\frac13 EB, EZ=\frac13 ED</math>, | ||
+ | |||
+ | <math>XY//GE</math>, since <math>BX=\frac23BE, BY=\frac23BG, EG//AD</math> by midsegment theorem, so <math>XY//AD</math> | ||
+ | |||
+ | Similarly, <math>YZ//AB</math>, | ||
+ | |||
+ | So <math>\triangle ABD</math> is an equilateral triangle | ||
+ | |||
+ | Assume <math>\alpha=\angle BCD</math>, then <math>BD^2=BC^2+CD^2-2\cdot BC\cdot CD\cos \alpha=40-24\cos\alpha</math>, the area | ||
+ | |||
+ | <math>[ABCD]=[ABD]+[BCD]=\frac{\sqrt3}4 BD^2+\frac12\cdot BC\cdot CD\sin\alpha=</math> | ||
+ | |||
+ | <math>10\sqrt3+6(\sin\alpha-\sqrt3\cos \alpha)=10\sqrt3+12\sin(\alpha-60^\circ)</math> | ||
+ | |||
+ | |||
+ | The maximal value happens when <math>\sin(\alpha-60^\circ)=1</math>, and the value is <math>10\sqrt3+12</math>, and the answer is <math>\boxed{\textbf{(C)} 12+10\sqrt3}</math>. | ||
+ | |||
+ | <asy> | ||
+ | import graph; size(11.42cm); | ||
+ | real labelscalefactor = 0.5; /* changes label-to-point distance */ | ||
+ | pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ | ||
+ | pen dotstyle = black; /* point style */ | ||
+ | real xmin = -1.58, xmax = 9.84, ymin = -7.74, ymax = 8.48; /* image dimensions */ | ||
+ | |||
+ | /* draw figures */ | ||
+ | draw((0.,0.)--(4.,0.), linewidth(2.)); | ||
+ | draw((2.,3.4641016151377544)--(4.,0.), linewidth(2.)); | ||
+ | draw((0.,0.)--(2.,3.4641016151377544), linewidth(2.)); | ||
+ | draw((4.,0.)--(5.,1.), linewidth(2.)); | ||
+ | draw((5.,1.)--(2.,3.4641016151377544), linewidth(2.)); | ||
+ | draw((0.,0.)--(5.,1.), linewidth(2.)); | ||
+ | draw((2.5,0.5)--(4.,0.), linewidth(2.)); | ||
+ | draw((4.,0.)--(3.5,2.232050807568877), linewidth(2.)); | ||
+ | draw((2.,3.4641016151377544)--(2.5,0.5), linewidth(2.)); | ||
+ | draw((0.,0.)--(3.5,2.232050807568877), linewidth(2.)); | ||
+ | draw((0.,0.)--(4.5,0.5), linewidth(2.)); | ||
+ | draw((2.,3.4641016151377544)--(4.5,0.5), linewidth(2.)); | ||
+ | draw((2.333333333333333,1.4880338717125845)--(3.,0.3333333333333333), linewidth(2.) + linetype("2 2")); | ||
+ | draw((3.666666666666666,1.488033871712585)--(3.,0.3333333333333333), linewidth(2.) + linetype("2 2")); | ||
+ | /* dots and labels */ | ||
+ | dot((0.,0.),dotstyle); | ||
+ | label("$A$", (-0.2,-0.18), NE * labelscalefactor); | ||
+ | dot((4.,0.),dotstyle); | ||
+ | label("$B$", (3.98,-0.3), NE * labelscalefactor); | ||
+ | dot((2.,3.4641016151377544),dotstyle); | ||
+ | label("$D$", (1.86,3.62), NE * labelscalefactor); | ||
+ | dot((5.,1.),dotstyle); | ||
+ | label("$C$", (5.06,0.92), NE * labelscalefactor); | ||
+ | dot((2.5,0.5),linewidth(4.pt) + dotstyle); | ||
+ | label("$E$", (2.2,0.5), NE * labelscalefactor); | ||
+ | dot((4.5,0.5),linewidth(4.pt) + dotstyle); | ||
+ | label("$F$", (4.52,0.3), NE * labelscalefactor); | ||
+ | dot((3.5,2.232050807568877),linewidth(4.pt) + dotstyle); | ||
+ | label("$G$", (3.58,2.4), NE * labelscalefactor); | ||
+ | dot((2.333333333333333,1.4880338717125845),linewidth(4.pt) + dotstyle); | ||
+ | label("$Z$", (2.06,1.44), NE * labelscalefactor); | ||
+ | dot((3.,0.3333333333333333),linewidth(4.pt) + dotstyle); | ||
+ | label("$X$", (2.92,0.04), NE * labelscalefactor); | ||
+ | dot((3.666666666666666,1.488033871712585),linewidth(4.pt) + dotstyle); | ||
+ | label("$Y$", (3.76,1.5), NE * labelscalefactor); | ||
+ | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | ||
+ | </asy> | ||
+ | |||
+ | |||
+ | ~szhangmath | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2019|ab=B|num-b=24|after=Last Problem}} | {{AMC12 box|year=2019|ab=B|num-b=24|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 02:38, 10 October 2024
Contents
Problem
Let be a convex quadrilateral with and Suppose that the centroids of and form the vertices of an equilateral triangle. What is the maximum possible value of the area of ?
Solution 1 (vectors)
Place an origin at , and assign position vectors of and . Since is not parallel to , vectors and are linearly independent, so we can write for some constants and . Now, recall that the centroid of a triangle has position vector .
Thus the centroid of is ; the centroid of is ; and the centroid of is .
Hence , , and . For to be equilateral, we need . Further, . Hence we have , so is equilateral.
Now let the side length of be , and let . By the Law of Cosines in , we have . Since is equilateral, its area is , while the area of is . Thus the total area of is , where in the last step we used the subtraction formula for . Alternatively, we can use calculus to find the local maximum. Observe that has maximum value when e.g. , which is a valid configuration, so the maximum area is .
Solution 2
Let , , be the centroids of , , and respectively, and let be the midpoint of . , , and are collinear due to well-known properties of the centroid. Likewise, , , and are collinear as well. Because (as is also well-known) and , we have . This implies that is parallel to , and in terms of lengths, . (SAS Similarity)
We can apply the same argument to the pair of triangles and , concluding that is parallel to and . Because (due to the triangle being equilateral), , and the pair of parallel lines preserve the angle, meaning . Therefore is equilateral.
At this point, we can finish as in Solution 1, or, to avoid using trigonometry, we can continue as follows:
Let , where due to the Triangle Inequality in . By breaking the quadrilateral into and , we can create an expression for the area of . We use the formula for the area of an equilateral triangle given its side length to find the area of and Heron's formula to find the area of .
After simplifying,
Substituting , the expression becomes
We can ignore the for now and focus on .
By the Cauchy-Schwarz inequality,
The RHS simplifies to , meaning the maximum value of is .
Thus the maximum possible area of is .
Solution 3 (Complex Numbers)
Let , , , and correspond to the complex numbers , , , and , respectively. Then, the complex representations of the centroids are , , and . The pairwise distances between the centroids are , , and , all equal. Thus, , so . Hence, is equilateral.
By the Law of Cosines, .
. Thus, the maximum possible area of is .
~ Leo.Euler
Solution 4 (Homothety)
Let , and be the centroids of , and , respectively, and let and be the midpoints of and , respectively. Note that and are of the way from to and , respectively, by a well-known property of centroids. Then a homothety centered at with ratio maps and to and , respectively, implying that is equilateral too. But is the medial triangle of , so is also equilateral. We may finish with the methods in the solutions above.
~ numberwhiz
While the solutions above have attempted the problem in general, knowing the fact that is equilateral greatly reduces the effort to find the final answer, hence I propose an alternative after this.
Let and . By cosine rule on : Thus, the total area of the quadrilateral is supposedly : Where the inequality comes from a common trigonometric identity,
~ SouradipClash_03
Video Solution by MOP 2024
~r00tsOfUnity
Solution 5
Let be the centroids of respectively, then
, since ,
, since by midsegment theorem, so
Similarly, ,
So is an equilateral triangle
Assume , then , the area
The maximal value happens when , and the value is , and the answer is .
~szhangmath
See Also
2019 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Problem |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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