View source for FidgetBoss 4000's 2019 Mock AMC 12B Problems/Problem 2

==Problem==
<i>(fidgetboss_4000)</i> In the diagram below, <math>ABC</math> is an isosceles right triangle with a right angle at <math>B</math> and with a hypotenuse of <math>40\sqrt2</math>  units. Find the greatest integer less than or equal to the value of the radius of the quarter circle inscribed inside <math>\triangle ABC</math>.

<math>\textbf{(A) } 26\qquad\textbf{(B) } 27\qquad\textbf{(C) } 28\qquad\textbf{(D) } 29\qquad\textbf{(E) } 30\qquad</math>

==Solution==
The quarter circle is centered at <math>B</math> and just touches the hypotenuse at the midpoint of the hypotenuse due to symmetry in an isosceles right triangle. The value of its radius is equal to the distance between <math>B</math> and the midpoint of the hypotenuse, and it is well known that this is half the length of the hypotenuse, or <math>\frac{1}{2}\cdot 40\sqrt2=20\sqrt2</math>. The greatest integer less than or equal to <math>20\sqrt2</math> is <math>28</math>, thus we pick answer <math>\boxed{\textbf{(C) }28}</math>.

==See also==
{{AMC12 box|year=FidgetBoss 4000's 2019 Mock|ab=B|before=[[FidgetBoss 4000's 2019 Mock AMC 12B Problems/Problem 1|Problem 1]]|after=[[FidgetBoss 4000's 2019 Mock AMC 12B Problems/Problem 3|Problem 3]]}}

[[Category:Introductory Geometry Problems]]
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