ISL 2001 G7

by Wolstenholme, Aug 2, 2014, 4:56 AM

Let $O$ be an interior point of acute triangle $ABC$ . Let $A_1$ lie on $BC$ with $OA_1$ perpendicular to $BC$ . Define $B_1$ on $CA$ and $C_1$ on $AB$ similarly. Prove that $O$ is the circumcenter of $ABC$ if and only if the perimeter of $A_1B_1C_1$ is not less than any one of the perimeters of $AB_1C_1, BC_1A_1$ , and $CA_1B_1$ .

Proof:

First note that if $O$ is the circumcenter of $\triangle{ABC}$ then the perimeters of all four relevant triangles are equal as desired.

Now, assume $O$ is not the circumcenter of $\triangle{ABC}$ . Let $R$ be the circumradius of $\triangle{ABC}$ and let $OC = \max{(OA, OB, OC)}$ . I shall show that the perimeter of $\triangle{A_{1}B_{1}C}$ is greater than that of $\triangle{A_{1}B_{1}C_{1}}$ .

It suffices to show that $A_{1}C + B_{1}C > B_{1}C_{1} + A_{1}C_{1} = \frac{OA \cdot BC}{2R} + \frac{OB \cdot AC}{2R}$ .

Now $A_{1}C = OC \cdot \cos{BCO} = \frac{OC^2 + BC^2 - OB^2}{2BC}$ and similarly $B_{1}C = \frac{OC^2 + AC^2 - OA^2}{2AC}$ so it suffices to show that $BC \cdot \frac{OA}{R} + AC \cdot \frac{OB}{R} < \frac{OC^2 - OB^2}{BC} + \frac{OC^2 - OA^2}{AC} + AC + BC$ .

Now by the Triangle Inequality we have that $OB + OC > BC$ and $OA + OC > AC$ so it suffices to show that $BC \cdot \frac{OA}{R} + AC \cdot \frac{OB}{R} < 2OC - OA - OB + AC + BC$ . Since the circles centered at $A, B, C$ with radius $R$ completely cover $\triangle{ABC}$ we have that $\min{(OA, OB, OC)} < R$ so we can assume WLOG that $OA < R$ . Then it suffices to show that $AC \cdot (\frac{OB}{R} - 1) \leq 2OC - OA - OB$ . But clearly $AC \cdot (\frac{OB}{R} - 1) \leq 2OB - 2R \leq 2OC - 2R$ so it suffices to show that $2R \geq OA + OB$ .

Now let the midpoints of segments $AC, BC$ be $M, N$ respectively. There exists a point $X$ such that $X$ is either on segment $AB$ or on segment $BC$ and $MX \perp AC$ . There exists a point $Y$ such that $Y$ is either on segment $AB$ or on segment $AC$ and $NY \perp BC$ . Since $OC = \max{(OA, OB, OC)}$ and since $O$ lies in the interior of $\triangle{ABC}$ , we have that $O$ lies in either triangle $XO'Y$ , quadrilateral $AXO'Y$ , quadrilateral $BXO'Y$ , or pentagon $ABXO'Y$ where $O' = MX \cap NY$ is the circumcenter of $\triangle{ABC}$ (it depends what segments $X$ and $Y$ are on).

By convexity, $OA + OB$ is maximized when $O$ is at $A, B, X$ , or $Y$ (since we assumed $O \ne O'$ ). If $O \in AB$ then $OA + OB = AB \leq 2R$ as desired.

So, it suffices to show that if $X \in BC$ then $XA + XB \leq 2R$ . But we have that $XA + XB \leq XB + XC = BC \leq 2R$ as desired so we are done.

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by mishka1980, Sep 12, 2015, 10:33 PM
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by donot, Aug 31, 2015, 11:07 PM
"I am Sam"
"Sam I am"
by mathwizard888, Aug 12, 2015, 9:13 PM
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by Eugenis, Apr 20, 2015, 10:10 PM
Uh-oh ARML practice is Thursday... I should start the homework.
by nosaj, Apr 20, 2015, 12:34 AM
Yes I am Sam, and Chebyshev polynomials aren't trivial, although they do make some problems trivial
by Wolstenholme, Apr 15, 2015, 10:00 PM
How are Chebyshev Polynomials trivial?
by nosaj, Apr 13, 2015, 4:10 AM
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by Eugenis, Apr 4, 2015, 2:05 AM
@Brian: yes, yes I did #whoneedsalgskillz?

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by Wolstenholme, Mar 1, 2015, 11:25 PM
hello!!!
by gauss1181, Nov 27, 2014, 12:19 AM
Hi Wolstenholme did you actually use calc on that tstst problem
by briantix, Aug 2, 2014, 12:25 AM

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ISL 2001 G7

by Wolstenholme, Aug 2, 2014, 4:56 AM

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