Difference between revisions of "1967 AHSME Problems/Problem 40"
(Created page with "== Problem == Located inside equilateral triangle <math>ABC</math> is a point <math>P</math> such that <math>PA=8</math>, <math>PB=6</math>, and <math>PC=10</math>. To the neare...") |
m (→Solution 4(Answer Choices, Approximation)) |
||
(14 intermediate revisions by 3 users not shown) | |||
Line 8: | Line 8: | ||
\textbf{(E)}\ 50</math> | \textbf{(E)}\ 50</math> | ||
− | == Solution == | + | == Solution 1 == |
− | <math>\ | + | |
+ | <asy> | ||
+ | draw((0,10)--(8.66,-5)--(-8.66,-5)--cycle); | ||
+ | label("$A$",(0,10),N); | ||
+ | label("$B$",(-9.5,-5.2),N); | ||
+ | label("$C$",(9.5,-5.2),N); | ||
+ | |||
+ | dot((-3,0)); | ||
+ | label("$P$",(-3,-2),N); | ||
+ | draw((-3,0)--(0,10)); | ||
+ | draw((-3,0)--(-8.66,-5)); | ||
+ | draw((-3,0)--(8.66,-5)); | ||
+ | |||
+ | dot((-9,7.5)); | ||
+ | label("$P'$",(-9.2,7.5),N); | ||
+ | draw((-9,7.5)--(0,10)); | ||
+ | draw((-9,7.5)--(-8.66,-5)); | ||
+ | draw((-9,7.5)--(-3,0)); | ||
+ | |||
+ | </asy> | ||
+ | |||
+ | Notice that <math>6^2+8^2=10^2.</math> That makes us want to construct a right 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> | ||
+ | |||
+ | Therefore, <math>\triangle APP'</math> is equilateral, so <math>P'P=8</math>, which means <math>\angle P'PB = 90^{\circ}.</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 APC</math> (remembering <math>\angle APC = \angle AP'B</math>): | ||
+ | <cmath> | ||
+ | |||
+ | 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 | ||
+ | |||
+ | == 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 14:48, 11 November 2023
Contents
[hide]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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.