<PRE><B>Question 1210173</B>
&lt;br&gt;
Also borrowing the diagram from Edwin, and modifying it some for my use....&lt;br&gt;
{{{drawing(400,1600/9,-1,8,-1,3,

triangle(0,0,7,0,2.357142857,1.855768722),

green(line(0,0,3.78882,1.28251)),

locate(0,0,B),locate(7,0,C), locate(2.27,2.18,A),
green(locate(3.77,1.6,E)),

locate(3,2,x),

locate(1.2,1.53,3), locate(3.5,0,7), locate(4.5,1.4,&quot;5-x&quot;),

green(locate(2.2,.8,4))


 )}}}
If point D is anywhere on AE between A and E (i.e., if point P is inside triangle ABE), the length of BD will be less than 4, as required.&lt;br&gt;
(1) Use Heron&#39;s formula to find the area of triangle ABC.&lt;br&gt;
{{{sqrt((15/2)(9/2)(5/2)(1/2))=(15sqrt(3))/4}}}&lt;br&gt;
(2) Use Stewart&#39;s Theorem to determine the length of AE.&lt;br&gt;
{{{(7)(x)(7)+(3)(5-x)(3)=(4)(5)(4)+(5)(x)(5-x)}}}
{{{49x+45-9x=80+25x-5x^2}}}
{{{5x^2+15x-35=0}}}
{{{x^2+3x-7=0}}}&lt;br&gt;
{{{x=(-3+sqrt(37))/2}}} (ignore the negative root!)&lt;br&gt;
(3) use the law of cosines to find the measure of angle A.&lt;br&gt;
{{{7^2=3^2+5^2-2(3)(5)cos(A)}}}
{{{49=34-30cos(A)}}}
{{{15=-30cos(A)}}}
{{{cos(A)=-.5}}}&lt;br&gt;
The measure of angle A is 120 degrees.&lt;br&gt;
(4) Find the area of triangle ABE using area = (1/2)(AB)(BE)sin(A).&lt;br&gt;
Cos(A) is 0.5, and angle A is obtuse, so sin(A) is {{{(-sqrt(3)/2)}}}&lt;br&gt;
{{{(1/2)(3)((-3+sqrt(37))/2)((-sqrt(3)/2))=(-9sqrt(3)+3sqrt(111))/8}}}&lt;br&gt;
(5) The probability in the question is the ratio of the areas of triangles ABE and ABC:&lt;br&gt;
{{{((-9sqrt(3)+3sqrt(111))/8)/((15sqrt(3))/4)}}}&lt;br&gt;
To several decimal places that number is 0.30828.&lt;br&gt;
ANSWER: The probability is approximately 0.30828, or about 30.8%&lt;br&gt;
That agrees with the answer Edwin got -- reached by a very different path.&lt;br&gt;
And now a final note on this problem....&lt;br&gt;
In Edwin&#39;s figure, vertex B is opposite the side with length 5, leading to the solution above.&lt;br&gt;
However, there is nothing in the statement of the problem -- either specific or implied -- that that is the case.&lt;br&gt;
And, indeed, it is easy to see that if B is the vertex opposite the side with length 3, the length of AD will NEVER be less than 4; and if B is opposite the side with length 7, the length of AD will ALWAYS be less than 4.&lt;br&gt;
Since we could choose (with equal probability) any of the vertices of triangle ABC to be angle B, perhaps the ANSWER to the problem should be the average of the probabilities for the three cases:&lt;br&gt;
{{{(0.308+0+1)/3=0.436}}}&lt;br&gt;
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