Difference between revisions of "1967 AHSME Problems/Problem 40"
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\textbf{(E)}\ 50</math> | \textbf{(E)}\ 50</math> | ||
− | == Solution == | + | == Solution 1 == |
<asy> | <asy> | ||
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Notice that <math>6^2+8^2=10^2.</math> That makes us want to construct a right triangle. | Notice that <math>6^2+8^2=10^2.</math> That makes us want to construct a right triangle. | ||
− | Rotate <math>\triangle | + | Rotate <math>\triangle APC</math> <math>60^{\circ}</math> about A. Note that <math>\triangle PAC \cong \triangle P'AB</math>, so |
<cmath>\angle P'AP = \angle PAB + \angle P'AB = \angle PAB + \angle PAC = 60^{\circ}.</cmath> | <cmath>\angle P'AP = \angle PAB + \angle P'AB = \angle PAB + \angle PAC = 60^{\circ}.</cmath> | ||
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Let <math>\angle BP'P = \alpha .</math> Notice that <math>\cos\alpha = \frac{8}{10}=\frac{4}{5},</math> and <math>\sin\alpha = \frac{3}{5}.</math> | Let <math>\angle BP'P = \alpha .</math> Notice that <math>\cos\alpha = \frac{8}{10}=\frac{4}{5},</math> and <math>\sin\alpha = \frac{3}{5}.</math> | ||
− | Applying the Law of Cosines to <math>\triangle | + | Applying the Law of Cosines to <math>\triangle APC</math> (remembering <math>\angle APC = \angle AP'B</math>): |
− | <cmath>\begin{align*} AC^2 &= 10^2+8^2-2\cdot10\cdot8\cdot \cos(60^{\circ}+\alpha)\\&= 164-160(\cos60\cos\alpha-\sin60\sin\alpha)\\&= | + | <cmath>\begin{align*} AC^2 &= 10^2+8^2-2\cdot10\cdot8\cdot \cos(60^{\circ}+\alpha)\\&= 164-160(\cos60\cos\alpha-\sin60\sin\alpha)\\&= 164-160(\frac{2}{5}-\frac{3\sqrt3}{10}) \\&= 164-16(4-3\sqrt3) \\ &= 100+48\sqrt3.\end{align*}</cmath> |
− | We want to find the area of <math>\triangle ABC</math> | + | We want to find the area of <math>\triangle ABC</math>, which is <cmath>AC^2\cdot\frac{\sqrt3}{4}=25\sqrt3+36\approx\boxed{(D) 79}.</cmath> |
~pfalcon | ~pfalcon | ||
+ | |||
+ | == Solution 2 (Magic Formula) == | ||
+ | |||
+ | Fun formula: Given a point whose distances from the vertices of an equilateral triangle are <math>a</math>, <math>b</math>, and <math>c</math>, the side length of the triangle is: | ||
+ | |||
+ | <cmath>s=\sqrt{\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)}</cmath> | ||
+ | |||
+ | Given that the area of an equilateral triangle is <math>\frac{\sqrt{3}}{4}s^2</math>, the answer is: | ||
+ | |||
+ | <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{96(144+400+225)-48(81+256+625)}\right)\\ | ||
+ | &= \frac{\sqrt{3}}{8}\left(200\pm\sqrt{96\cdot769-48\cdot962}\right) = \frac{\sqrt{3}}{8}\left(200\pm\sqrt{96\cdot769-96\cdot481}\right)\\ | ||
+ | &= \frac{\sqrt{3}}{8}\left(200\pm\sqrt{96\cdot288}\right) = \frac{\sqrt{3}}{8}\left(200\pm\sqrt{96\cdot96\cdot3}\right)\\ | ||
+ | &= 25\sqrt{3}\pm36 \approx \{6.5, \text{or } 78.5\} | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | <math>6.5</math> is not a choice, therefore the answer is <math>\boxed{\textbf{(D) }79}</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 | ||
+ | |||
+ | ==Solution 4(Answer Choices, Approximation)== | ||
+ | Let <math>s</math> be the side length of <math>ABC.</math> Notice that <math>s\le 14</math> by the triangle inequality. This means that | ||
+ | <cmath>[ABC]\le\dfrac{14^2\sqrt{3}}{2}\approx 84.87.</cmath> This automatically rules out choices <math>A, B,</math> and <math>C.</math> Now, we will look at if the area is <math>50</math>. By the equilateral triangle area formula, <math>s</math> would equal <math>10\sqrt{\dfrac{2}{\sqrt{3}}}\approx 10.75.</math> This is very close to <math>10.</math> If <math>s=10,</math> <math>\angle APB=90</math> and <math>\angle APC, \angle BPC<90</math> by the Pythagorean theorem and Pythagorean inequalities. Thus, <cmath>\angle APB+\angle APC+\angle BPC<270.</cmath> | ||
+ | <math>\angle APB+\angle APC+\angle BPC</math> needs to be <math>360,</math> and it probably cannot increase by more than <math>90</math> by just adding <math>0.75</math> to <math>s</math> (more rigorous proof below) Thus, the only viable answer choice is <math>\boxed{79}.</math> | ||
+ | |||
+ | *In fact, for <math>s<\sqrt{136}\approx 11.66</math> (<math>[ABC]<34\sqrt{3}\approx58.88</math>) <math>\angle APC, \angle BPC<90</math> still holds. As <math>\angle APB <180,</math> <cmath>\angle APB+\angle APC+\angle BPC<360.</cmath> Thus, we know for sure that the nearest integer to the area cannot be <math>50.</math> | ||
== See also == | == See also == | ||
− | {{AHSME box|year=1967|num-b=39| | + | {{AHSME 40p box|year=1967|num-b=39|after=Last Problem}} |
[[Category: Intermediate Geometry Problems]] | [[Category: Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 15:48, 11 November 2023
Contents
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
Solution 4(Answer Choices, Approximation)
Let be the side length of
Notice that
by the triangle inequality. This means that
This automatically rules out choices
and
Now, we will look at if the area is
. By the equilateral triangle area formula,
would equal
This is very close to
If
and
by the Pythagorean theorem and Pythagorean inequalities. Thus,
needs to be
and it probably cannot increase by more than
by just adding
to
(more rigorous proof below) Thus, the only viable answer choice is
- In fact, for
(
)
still holds. As
Thus, we know for sure that the nearest integer to the area cannot be
See also
1967 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 39 |
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
All AHSME Problems and Solutions |
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