document.write( "Question 1210173: Let P be a point chosen uniformly at random inside triangle ABC. Extend ray BP to hit side AC at D. What is the probability that BD < 4?\r
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\n" ); document.write( "\n" ); document.write( "The sides of triangle ABC are 3, 5, and 7.
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Algebra.Com's Answer #851447 by CPhill(1959)\"\" \"About 
You can put this solution on YOUR website!
Let's break down this problem step by step.\r
\n" ); document.write( "\n" ); document.write( "**1. Geometric Interpretation**\r
\n" ); document.write( "\n" ); document.write( "We are looking for the probability that BD < 4. This is equivalent to finding the area of the region inside triangle ABC where BD < 4, and dividing it by the total area of triangle ABC.\r
\n" ); document.write( "\n" ); document.write( "**2. Triangle ABC and Ray BP**\r
\n" ); document.write( "\n" ); document.write( "* Let P be a random point inside triangle ABC.
\n" ); document.write( "* Ray BP intersects AC at point D.
\n" ); document.write( "* We want to find the probability that BD < 4.\r
\n" ); document.write( "\n" ); document.write( "**3. Law of Cosines**\r
\n" ); document.write( "\n" ); document.write( "Let's find the angles of triangle ABC using the Law of Cosines:\r
\n" ); document.write( "\n" ); document.write( "* Let a = 3, b = 5, c = 7.
\n" ); document.write( "* $\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{25 + 49 - 9}{2(5)(7)} = \frac{65}{70} = \frac{13}{14}$
\n" ); document.write( "* $\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{9 + 49 - 25}{2(3)(7)} = \frac{33}{42} = \frac{11}{14}$
\n" ); document.write( "* $\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{9 + 25 - 49}{2(3)(5)} = \frac{-15}{30} = -\frac{1}{2}$\r
\n" ); document.write( "\n" ); document.write( "Since $\cos C = -1/2$, angle C = 120 degrees.\r
\n" ); document.write( "\n" ); document.write( "**4. Area of Triangle ABC**\r
\n" ); document.write( "\n" ); document.write( "We can use Heron's formula to find the area of triangle ABC.\r
\n" ); document.write( "\n" ); document.write( "* Semi-perimeter: $s = (3 + 5 + 7)/2 = 15/2 = 7.5$
\n" ); document.write( "* Area = $\sqrt{s(s-a)(s-b)(s-c)} = \sqrt{7.5(4.5)(2.5)(0.5)} = \sqrt{42.1875} = \frac{15\sqrt{3}}{4}$\r
\n" ); document.write( "\n" ); document.write( "**5. Geometric Reasoning**\r
\n" ); document.write( "\n" ); document.write( "* Let's consider the locus of points P inside triangle ABC such that BD = 4.
\n" ); document.write( "* We can create a triangle ABD' where BD' = 4.
\n" ); document.write( "* The locus of P such that BD = 4 is a line parallel to AD'.
\n" ); document.write( "* The region where BD < 4 is the region inside triangle ABC that is closer to B than to D'.\r
\n" ); document.write( "\n" ); document.write( "**6. Ratio of Areas**\r
\n" ); document.write( "\n" ); document.write( "* The probability we are looking for is the ratio of the area of the region where BD < 4 to the area of the triangle ABC.
\n" ); document.write( "* Let's consider a point P that makes BD = 4.
\n" ); document.write( "* Let's determine the ratio of AD/CD.
\n" ); document.write( "* By the Angle Bisector Theorem, if P were to lie on an angle bisector, we could find the ratios. However, P is not necessarily on an angle bisector.
\n" ); document.write( "* The ratio AD/CD is dependent on the position of point P.\r
\n" ); document.write( "\n" ); document.write( "**7. Difficulties**\r
\n" ); document.write( "\n" ); document.write( "* Determining the exact shape of the region where BD < 4 is challenging without further geometric constructions and calculations.
\n" ); document.write( "* The problem requires a more advanced approach, potentially involving barycentric coordinates or other geometric methods.\r
\n" ); document.write( "\n" ); document.write( "**8. Conclusion**\r
\n" ); document.write( "\n" ); document.write( "Without further information or a more advanced approach, it's difficult to find the exact probability. However, we know that the probability is the ratio of the area of the region where BD < 4 to the total area of the triangle.\r
\n" ); document.write( "\n" ); document.write( "Unfortunately, I don't see a way to solve this problem without more advanced techniques.
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