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11 Jan 2019, 07:28
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MathRevolution
\(\left( {r,s} \right)\,\,\,\mathop \in \limits^? \,\,\,\left\{ {\,\left( {x,y} \right)\,\,:\,\,{{\left( {x - 0} \right)}^2} + {{\left( {y - 0} \right)}^2} \leqslant {5^2}\,} \right\}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\boxed{\,\,\,{r^2} + {s^2}\,\,\,\mathop \leqslant \limits^? \,\,\,25\,}\,\,\)
\(\left( 1 \right)\,\,\,\,\left| r \right| < 3\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,\left( {r,s} \right) = \left( {0,0} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr \\
\,{\rm{Take}}\,\,\left( {r,s} \right) = \left( {0,6} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.\)
\(\left( 2 \right)\,\,\,\left| s \right| < 4\,\,\,\,\left\{ \matrix{\\
\,\left( {{\mathop{\rm Re}\nolimits} } \right){\rm{Take}}\,\,\left( {r,s} \right) = \left( {0,0} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr \\
\,{\rm{Take}}\,\,\left( {r,s} \right) = \left( {6,0} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.\)
\(\left( {1 + 2} \right)\,\,\,\,\left\{ \matrix{\\
\,{r^2} = {\left| r \right|^2} < {3^2} \hfill \cr \\
\,{s^2} = {\left| s \right|^2} < {4^2} \hfill \cr} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{r^2} + {s^2} < 25\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle\)
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
11 Jan 2019, 08:09
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MathRevolution
\(x + y + z\,\,\mathop > \limits^? \,\,\,1\)
\(\left( 1 \right)\,\,\,\left\{ \matrix{\\
\,x\mathop > \limits^{\left( * \right)} \,\,\,h = {1 \over 2} \hfill \cr \\
\,z\mathop > \limits^{\left( * \right)} \,\,\,h = {1 \over 2} \hfill \cr \\
y > 0 \hfill \cr} \right.\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,x + y + z > 1\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle\)
\(\left( 2 \right)\,\,\left\{ \matrix{\\
\,{\rm{figure}}\,\,{\rm{on}}\,\,{\rm{the}}\,\,{\rm{left}}\,\,\, \Rightarrow \,\,\,\,\,x + y + z\,\,\,\, < \,\,\,\,3\left( {{1 \over 3}} \right)\,\, = \,\,1\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \hfill \cr \\
\,{\rm{figure}}\,\,{\rm{on}}\,\,{\rm{the}}\,\,{\rm{right}}\,\,\, \Rightarrow \,\,\,\,\,x + y + z\,\,\,\,\mathop < \limits^{{\rm{as}}\,{\rm{near}}\,\,{\rm{as}}\,\,{\rm{desired}}!} \,\,\,\,{2 \over 3} + {1 \over 3}\left( {\sqrt 2 } \right)\,\, = \,\,{{\sqrt 2 + 2} \over 3}\,\,\,\,\,\left[ { > \,\,\,1} \right]\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \hfill \cr} \right.\)
The correct answer is therefore (A).
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
P.S.: it is interesting to study the case in which D does not belong to the side BC!
13 Jan 2019, 17:52
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MathRevolution
=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.
In a triangle, the sum of the lengths of two sides is always greater than the length of the other side.
Thus, from triangle \(ABD\), we must have \(AB + BD > AD = h\), and from triangle \(ABD\), we must have \(AC + CD > AD = h\). If \(h = \frac{1}{2}\), then \(AB + BC + CA = AB + BD + DC + CA > \frac{1}{2} + \frac{1}{2} = 1.\)
Condition 1) is sufficient.
Condition 2):
If \(AC = \frac{1}{2}\), then \(AB + BC + CA = \frac{1}{3} + \frac{1}{3} + \frac{1}{2} = \frac{7}{6} > 1\) and the answer is ‘yes’.
If \(AC = \frac{1}{4}\), then \(AB + BC + CA = \frac{1}{3} + \frac{1}{3} + \frac{1}{4} = \frac{11}{12} < 1\) and the answer is ‘no’.
Since it does not yield a unique solution, condition 2) is not sufficient.
Therefore, the correct answer is A.
Answer: A
