Difference between revisions of "2019 AMC 12B Problems/Problem 25"
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~ numberwhiz | ~ numberwhiz | ||
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+ | 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. | ||
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+ | 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== | ==Video Solution by MOP 2024== | ||
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===Solution 5=== | ===Solution 5=== | ||
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Let <math>X, Y, Z</math> be the centroids of <math>\triangle ABC, \triangle BCD, \triangle ACD</math> respectively, then | Let <math>X, Y, Z</math> be the centroids of <math>\triangle ABC, \triangle BCD, \triangle ACD</math> respectively, then | ||
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− | \ | + | <math>XZ//BD</math>, since <math>EX=\frac13 EB, EZ=\frac13 ED</math>, |
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+ | <math>XY//GE</math>, since <math>BX=\frac23BE, BY=\frac23BG, EG//AD</math> by midsegment theorem, so <math>XY//AD</math> | ||
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+ | Similarly, <math>YZ//AB</math>, | ||
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+ | So <math>\triangle ABD</math> is an equilateral triangle | ||
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+ | 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 | ||
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+ | <math>[ABCD]=[ABD]+[BCD]=\frac{\sqrt3}4 BD^2+\frac12\cdot BC\cdot CD\sin\alpha=</math> | ||
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+ | <math>10\sqrt3+6(\sin\alpha-\sqrt3\cos \alpha)=10\sqrt3+12\sin(\alpha-60^\circ)</math> | ||
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+ | 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>. | ||
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<asy> | <asy> | ||
import graph; size(11.42cm); | import graph; size(11.42cm); |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.