Please act as a match teacher and solve the math problem step by step.
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# Question:
How many 3-letter words can we make from the letters A, B, C, and D, if we are allowed to repeat letters, and we must use the letter A at least once? (Here, a word is an arbitrary sequence of letters.)
# Reasoning:
There are $4^3$ three letter words from A, B, C, and D, and there are $3^3$ three letter words from just B, C, and D. There must, then, be $4^3 - 3^3=64-27 = \\boxed{37}$ words from A, B, C, and D containing at least one A.
# Answer:
The answer is: \\boxed{37}
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# Question:
In the diagram, square $ABCD$ has sides of length $4,$ and $\\triangle ABE$ is equilateral. Line segments $BE$ and $AC$ intersect at $P.$ Point $Q$ is on $BC$ so that $PQ$ is perpendicular  to $BC$ and $PQ=x.$ [asy]\npair A, B, C, D, E, P, Q;\nA=(0,0);\nB=(4,0);\nC=(4,-4);\nD=(0,-4);\nE=(2,-3.464);\nP=(2.535,-2.535);\nQ=(4,-2.535);\ndraw(A--B--C--D--A--E--B);\ndraw(A--C);\ndraw(P--Q, dashed);\nlabel(\"A\", A, NW);\nlabel(\"B\", B, NE);\nlabel(\"C\", C, SE);\nlabel(\"D\", D, SW);\nlabel(\"E\", E, S);\nlabel(\"P\", P, W);\nlabel(\"Q\", Q, dir(0));\nlabel(\"$x$\", (P+Q)/2, N);\nlabel(\"4\", (A+B)/2, N);\n[/asy] Determine the measure of angle $BPC.$
# Reasoning:
Since $\\triangle ABE$ is equilateral, we know that $\\angle ABE=60^\\circ.$ Therefore, \\begin{align*}\n\\angle PBC &= \\angle ABC - \\angle ABE \\\\\n&= 90^\\circ-60^\\circ \\\\\n&=30^\\circ.\n\\end{align*} Since $AB=BC,$ we know that $\\triangle ABC$ is a right isosceles triangle and $$\\angle BAC=\\angle BCA=45^\\circ.$$ Then, $\\angle BCP =\\angle BCA=45^\\circ$ and \\begin{align*}\n\\angle BPC &= 180^\\circ-\\angle PBC - \\angle BCP \\\\\n&= 180^\\circ - 30^\\circ - 45^\\circ \\\\\n&=\\boxed{105^\\circ}.\n\\end{align*}.
# Answer:
The answer is: \\boxed{105}
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# Question:
Find the $\\emph{positive}$ real number(s) $x$ such that $\\frac{1}{2}\\left( 3x^2-1\\right) = \\left( x^2-50x-10\\right)\\left( x^2+25x+5\\right)$.
# Reasoning:
Write $a = x^2-50x-10$ and $b = x^2+25x+5$.  Then the equation given becomes\n\\[\\frac{a+2b-1}{2} = ab,\\]so $0=2ab-a-2b+1=(a-1)(2b-1)$. Then $a-1=x^2-50x-11=0$ or $2b-1=2x^2+50x+9=0$. The former has a positive root, $x=\\boxed{25 + 2\\sqrt{159}}$, while the latter does not.
# Answer:
The answer is: \\boxed{25 + 2\\sqrt{159}}
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# Question:
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# Reasoning:

