Difference between revisions of "1967 AHSME Problems/Problem 40"
(Added a solution using a magic formula.) |
(Added another solution using rotations and area.) |
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<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | + | [ABC] &= \frac{\sqrt{3}}{4}\cdot\frac{1}{2}\left(a^2+b^2+c^2\pm\sqrt{6(a^2b^2+b^2c^2+c^2a^2)-3(a^4+b^4+c^4)}\right)\ | |
&= \frac{\sqrt{3}}{8}\left(200\pm\sqrt{6\cdot 16(3^2 4^2+4^2 5^2+5^2 3^2)-3\cdot16(3^4+4^4+5^4)}\right)\ | &= \frac{\sqrt{3}}{8}\left(200\pm\sqrt{6\cdot 16(3^2 4^2+4^2 5^2+5^2 3^2)-3\cdot16(3^4+4^4+5^4)}\right)\ | ||
&= \frac{\sqrt{3}}{8}\left(200\pm\sqrt{96(144+400+225)-48(81+256+625)}\right)\ | &= \frac{\sqrt{3}}{8}\left(200\pm\sqrt{96(144+400+225)-48(81+256+625)}\right)\ | ||
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(Note that the <math>6.5</math> answer is actually the solution for when point <math>P</math> is ''exterior'' to <math>\triangle ABC</math>.) | (Note that the <math>6.5</math> answer is actually the solution for when point <math>P</math> is ''exterior'' to <math>\triangle ABC</math>.) | ||
+ | |||
+ | ~proloto | ||
+ | |||
+ | == Solution 3 == | ||
+ | |||
+ | Rotate <math>P</math> and <math>B</math> <math>60^{\circ}</math> CCW around <math>A</math>, becoming <math>X</math> and <math>C</math>. Rotate <math>P</math> and <math>C</math> <math>60^{\circ}</math> CCW around <math>B</math>, becoming <math>Y</math> and <math>A</math>. Rotate <math>P</math> and <math>A</math> <math>60^{\circ}</math> CCW around <math>C</math>, becoming <math>Z</math> and <math>B</math>: | ||
+ | |||
+ | <asy> | ||
+ | import graph; | ||
+ | import geometry; | ||
+ | size(12cm); | ||
+ | |||
+ | pair A, B, C, P, X, Y, Z; | ||
+ | |||
+ | // Define the equilateral triangle ABC | ||
+ | real a = sqrt(100+48*sqrt(3)); | ||
+ | A = (0, 0); | ||
+ | B = rotate(60)*A + (a, 0); | ||
+ | C = rotate(120)*B + (a, 0); | ||
+ | |||
+ | // Define point P using given distances | ||
+ | pair[] P_candidates = intersectionpoints(Circle(A,8), Circle(B,6)); | ||
+ | for (pair candidate : P_candidates) { | ||
+ | if (length(candidate - C) < 10.1 && length(candidate - C) > 9.9) { | ||
+ | P = candidate; | ||
+ | break; | ||
+ | } | ||
+ | } | ||
+ | |||
+ | // Rotate C and P about A through 60 degrees to get B and X | ||
+ | X = rotate(60,A)*P; | ||
+ | |||
+ | // Rotate A and P about B through 60 degrees to get C and Y | ||
+ | Y = rotate(60,B)*P; | ||
+ | |||
+ | // Rotate B and P about C through 60 degrees to get A and Z | ||
+ | Z = rotate(60,C)*P; | ||
+ | |||
+ | // Draw the triangle and the segments | ||
+ | draw(A--B--C--cycle); | ||
+ | draw(A--P); | ||
+ | draw(B--P); | ||
+ | draw(C--P); | ||
+ | |||
+ | // Connect X, Y, Z to P and to the vertices of the triangle | ||
+ | draw(X--P, dashed); | ||
+ | draw(Y--P, dashed); | ||
+ | draw(Z--P, dashed); | ||
+ | draw(X--A, dashed); | ||
+ | draw(X--C, dashed); | ||
+ | draw(Y--A, dashed); | ||
+ | draw(Y--B, dashed); | ||
+ | draw(Z--B, dashed); | ||
+ | draw(Z--C, dashed); | ||
+ | |||
+ | // Label the points | ||
+ | label("$A$", A, SW); | ||
+ | label("$B$", B, SE); | ||
+ | label("$C$", C, N); | ||
+ | label("$P$", P, NNE*2); | ||
+ | label("$X$", X, NW); | ||
+ | label("$Y$", Y, S); | ||
+ | label("$Z$", Z, E); | ||
+ | |||
+ | // Add the distances | ||
+ | label("$8$", (A+P)/2, NW); | ||
+ | label("$6$", (B+P)/2, NE); | ||
+ | label("$10$", (C+P)/2, N); | ||
+ | |||
+ | // Add right angle marks | ||
+ | draw(rightanglemark(C,X,P,15)); | ||
+ | draw(rightanglemark(P,B,Z,15)); | ||
+ | draw(rightanglemark(A,P,Y,15)); | ||
+ | </asy> | ||
+ | |||
+ | Notice that since <math>\triangle AXC\cong\triangle APB</math>, <math>\triangle BYA\cong\triangle BPC</math>, and <math>\triangle CZB\cong\triangle CPA</math>, then | ||
+ | |||
+ | <cmath>[AYBZCX]=2\cdot[ABC]</cmath> | ||
+ | |||
+ | Now the area of the big hexagon is easy to compute since it's comprised of 3 equilateral triangle and 3 right triangles: | ||
+ | |||
+ | <cmath>\begin{align*}[ABC] &= \frac{1}{2}[AYBZCX] = \frac{1}{2}\left(\underbrace{3\cdot\frac12\cdot6\cdot8}_{\text{3 right triangles}}+\underbrace{\frac{\sqrt{3}}{4}\left(6^2+8^2+10^2\right)}_{\text{3 equilateral triangles}}\right)\ | ||
+ | &= \frac{1}{2}\left(72+\frac{\sqrt{3}}{4}\cdot200)\right) = 36+25\sqrt{3}\ | ||
+ | &\approx \boxed{\textbf{(D) }79} | ||
+ | \end{align*}</cmath> | ||
~proloto | ~proloto |
Revision as of 00:24, 16 August 2023
Problem
Located inside equilateral triangle is a point such that , , and . To the nearest integer the area of triangle is:
Solution 1
Notice that That makes us want to construct a right triangle.
Rotate about A. Note that , so
Therefore, is equilateral, so , which means
Let Notice that and
Applying the Law of Cosines to (remembering ):
We want to find the area of , which is
~pfalcon
Solution 2 (Magic Formula)
Fun formula: Given a point whose distances from the vertices of an equilateral triangle are , , and , the side length of the triangle is:
Given that the area of an equilateral triangle is , the answer is:
is not a choice, therefore the answer is .
(Note that the answer is actually the solution for when point is exterior to .)
~proloto
Solution 3
Rotate and CCW around , becoming and . Rotate and CCW around , becoming and . Rotate and CCW around , becoming and :
Notice that since , , and , then
Now the area of the big hexagon is easy to compute since it's comprised of 3 equilateral triangle and 3 right triangles:
~proloto
See also
1967 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 39 |
Followed by Problem 40 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.