Difference between revisions of "1967 AHSME Problems/Problem 40"
m (Fixed a couple typos) |
(Added a solution using a magic formula.) |
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\textbf{(E)}\ 50</math> | \textbf{(E)}\ 50</math> | ||
− | == Solution == | + | == Solution 1 == |
<asy> | <asy> | ||
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~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*} | ||
+ | A &= \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 | ||
== See also == | == See also == |
Revision as of 23:20, 15 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
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.