fall-2017-final

Solutions to Spring 2017 HEXA Challenge Problems

April Challenge: Solve x log(5x) = 40 algebraically.

SOLUTION: (log(x)) stands for the logarithm of x. Take the logarithm of both sides: log(x log(5·x) ) = log(40) The left-hand side is equal to:

log(5·x) · log(x) = log((10/2) ∙ x) ∙ log(x) = (1 - log(2) + log(x)) ∙ log(x)

Respectively, the right hand side is:

log(10 ∙ 4) = log(10 ∙ 2 2 ) = log(10) + log(2 2 ) = 1 + 2 ∙ log(2)

Therefore:

(1 - log(2) + log(x)) ∙ log(x) = 1 + 2 ∙ log(2)

Let’s say z = log(x). Then:

(1 - log(2) + z) ∙ z = 1 + 2 ∙ log(2)

This reduces to a quadratic equation

z 2 + (1 - log(2)) ∙ z = 1 + 2 ∙ log(2)

whose roots are

z 1

=

,

z 2

=

where z 1

= log(x) and z 2

= log(x).

≈ 0.056; x 2

≈ 3.588

Hence the roots of the primarily suggested equation are: x 1

May Challenge: Construct a right triangle ABC where AC is the hypotenuse. Create a point D, such that BD is the height, per- pendicular to AC. In ∆ABD and ∆BCD, there are inscribed circles with radii r 1 and r 2 , respectively. Find the radius of the circle inscribed in ∆ABC.

SOLUTION: There are three similar triangles: ∆ADB ~ ∆BDC ~ ∆ABC. Therefore,

Virginia Mathematics Teacher vol. 44, no. 1

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