Difference between revisions of "2018 AMC 12B Problems/Problem 13"
MRENTHUSIASM (talk | contribs) (The original Solutions 2 and 3 are very similar and should be combined. I will combine them into Sol 1, as it is the most educational. Credits are retained to the original authors) |
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==Solution 1 (Similar Triangles and Area Ratios)== | ==Solution 1 (Similar Triangles and Area Ratios)== | ||
− | < | + | As shown below, let <math>M_1,M_2,M_3,M_4</math> be the midpoints of <math>\overline{AB},\overline{BC},\overline{CD},\overline{DA},</math> respectively, and <math>G_1,G_2,G_3,G_4</math> be the centroids of <math>\triangle{ABP},\triangle{BCP},\triangle{CDP},\triangle{DAP},</math> respectively. |
+ | <asy> | ||
+ | unitsize(210); | ||
+ | pair B = (0, 0), A = (0, 1), D = (1, 1), C = (1, 0), P = (1/4, 2/3); | ||
+ | pair M1 = midpoint(A--B); | ||
+ | pair M2 = midpoint(B--C); | ||
+ | pair M3 = midpoint(C--D); | ||
+ | pair M4 = midpoint(D--A); | ||
+ | pair G1 = centroid(A,B,P); | ||
+ | pair G2 = centroid(B,C,P); | ||
+ | pair G3 = centroid(C,D,P); | ||
+ | pair G4 = centroid(D,A,P); | ||
+ | filldraw(M1--M2--P--cycle,red); | ||
+ | filldraw(M2--M3--P--cycle,yellow); | ||
+ | filldraw(M3--M4--P--cycle,green); | ||
+ | filldraw(M4--M1--P--cycle,lightblue); | ||
+ | draw(A--B--C--D--cycle); | ||
+ | draw(M1--M2--M3--M4--cycle); | ||
+ | draw(G1--G2--G3--G4--cycle); | ||
+ | dot(P); | ||
+ | defaultpen(fontsize(10pt)); | ||
+ | draw(A--P--B); | ||
+ | draw(C--P--D); | ||
+ | label("$A$", A, W); | ||
+ | label("$B$", B, W); | ||
+ | label("$C$", C, E); | ||
+ | label("$D$", D, E); | ||
+ | label("$P$", P, N); | ||
+ | label("$M_1$", M1, W); | ||
+ | label("$M_2$", M2, S); | ||
+ | label("$M_3$", M3, E); | ||
+ | label("$M_4$", M4, N); | ||
+ | label("$G_1$", G1, 1.5S); | ||
+ | label("$G_2$", G2, 1.5E); | ||
+ | label("$G_3$", G3, 1.5NE); | ||
+ | label("$G_4$", G4, 1.5E); | ||
+ | dot(A); | ||
+ | dot(B); | ||
+ | dot(C); | ||
+ | dot(D); | ||
+ | dot(M1); | ||
+ | dot(M2); | ||
+ | dot(M3); | ||
+ | dot(M4); | ||
+ | dot(G1); | ||
+ | dot(G2); | ||
+ | dot(G3); | ||
+ | dot(G4); | ||
+ | </asy> | ||
+ | By SAS, we conclude that <math>\triangle G_1G_2P\sim\triangle M_1M_2P, \triangle G_2G_3P\sim\triangle M_2M_3P, \triangle G_3G_4P\sim\triangle M_3M_4P,</math> and <math>\triangle G_4G_1P\sim\triangle M_4M_1P.</math> By the properties of centroids, the side-length ratio for each pair of triangles is <math>\frac23,</math> so the area ratio for each pair of triangles is <math>\left(\frac23\right)^2=\frac49.</math> | ||
+ | Let the brackets denote areas. Since <math>[ABCD]=30^2=900,</math> it follows that | ||
+ | <cmath>\begin{align*} | ||
+ | [G_1G_2G_3G_4]&=[G_1G_2P]+[G_2G_3P]+[G_3G_4P]+[G_4G_1P] \ | ||
+ | &=\frac49[M_1M_2P]+\frac49[M_2M_3P]+\frac49[M_3M_4P]+\frac49[M_4M_1P] \ | ||
+ | &=\frac49([M_1M_2P]+[M_2M_3P]+[M_3M_4P]+[M_4M_1P]) \ | ||
+ | &=\frac49[M_1M_2M_3M_4] \ | ||
+ | &=\frac29[ABCD] | ||
+ | &=\boxed{\textbf{(C) }200}. | ||
+ | \end{align*}</cmath> | ||
~RandomPieKevin ~Kyriegon ~MRENTHUSIASM | ~RandomPieKevin ~Kyriegon ~MRENTHUSIASM | ||
Revision as of 15:20, 24 September 2021
Contents
[hide]Problem
Square has side length . Point lies inside the square so that and . The centroids of , , , and are the vertices of a convex quadrilateral. What is the area of that quadrilateral?
Solution 1 (Similar Triangles and Area Ratios)
As shown below, let be the midpoints of respectively, and be the centroids of respectively. By SAS, we conclude that and By the properties of centroids, the side-length ratio for each pair of triangles is so the area ratio for each pair of triangles is
Let the brackets denote areas. Since it follows that ~RandomPieKevin ~Kyriegon ~MRENTHUSIASM
Solution 2 (Coordinate Geometry)
We put the diagram on a coordinate plane. The coordinates of the square are and the coordinates of point P are By using the centroid formula, we find that the coordinates of the centroids are and Shifting the coordinates down by does not change its area, and we ultimately get that the area is equal to the area covered by which has an area of
Solution 3 (Accurate Diagram)
We can draw an accurate diagram by using centimeters and scaling everything down by a factor of The centroid is the intersection of the three medians in a triangle.
After connecting the centroids, we see that the quadrilateral looks like a square with side length of However, we scaled everything down by a factor of so the length is The area of a square is so the area is
Video Solution (Meta-Solving Technique)
https://youtu.be/GmUWIXXf_uk?t=1439
~ pi_is_3.14
See Also
2018 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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.