Geometry Question 1

Level pending

In the figure above, $r_{ 1 }$ , $r_{ 2 }$ , $r_{ 3 }$ are the radius of circles $O_{ 1 }$ , $O_{ 2 }$ , $O_{ 3 }$ respectively. Given that $O_{ 1 } + O_{ 2 }$ = $2 \times O_{ 3 }$ , and $r_{ 1 } = r_{ 2 }$ , find Segment $O_{ 3 }y$ in terms of $r_{ 1 }$ and $r_{ 3 }$ .

\sqrt{2}(r_{ 1 } + r_{ 3 })

r_{ 3 } + r_{ 1 }

\frac{\sqrt{2}( r_{ 1 } + r_{ 3 })}{2} + r_{ 1 }

\sqrt{\frac{r_{ 1 } + r_{ 3 }}{2}} + r_{ 3 }

\frac{\sqrt{2}(r_{ 1 } + r_{ 3 })}{2}

1 solution

Giwon Kim
Mar 14, 2015

By drawing a line on: $O_{ 1 }O_{ 3 }$ You can simply find $O_{ 3 }x$ using the Pythagorean Theorem:
$(r_{ 1 } + r_{ 3 })^{ 2 } = O_{ 3 }x^ { 2 } + O_{ 1 }x^ { 2 }$ $(r_{ 1 } + r_{ 3 })^{ 2 } = O_{ 3 }x^ { 2 } + O_{ 3 }x^ { 2 }$ $(r_{ 1 } + r_{ 3 })^{ 2 } = 2 \times O_{ 3 }x^ { 2 }$ $(r_{ 1 } + r_{ 3 }) = \sqrt{2} \times O_{ 3 }x$ $\frac{\sqrt{2} (r_{ 1 } + r_{ 3 })}{2} = O_{ 3 }x$ And Since: $O_{3}y = O_{3}x + r_{1}$ $\boxed{O_{3}y = \frac{\sqrt{2} (r_{ 1 } + r_{ 3 })}{2} + r_{1}}$

Geometry Question 1

1 solution

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