2001 Pan African MO Problems/Problem 3: Difference between revisions

Revision as of 00:15, 21 December 2019

Problem

Let $ABC$ be an equilateral triangle and let $P_0$ be a point outside this triangle, such that $\triangle{AP_0C}$ is an isosceles triangle with a right angle at $P_0$ . A grasshopper starts from $P_0$ and turns around the triangle as follows. From $P_0$ the grasshopper jumps to $P_1$ , which is the symmetric point of $P_0$ with respect to $A$ . From $P_1$ , the grasshopper jumps to $P_2$ , which is the symmetric point of $P_1$ with respect to $B$ . Then the grasshopper jumps to $P_3$ which is the symmetric point of $P_2$ with respect to $C$ , and so on. Compare the distance $P_0P_1$ and $P_0P_n$ . $n \in N$ .

Solution

We can use coordinate geometry to solve the problem. Let $P_0 = (0,0)$ , $A = (0,a)$ , and $C = (a,0)$ , making $AC = a \sqrt{2}$ . To calculate the coordinates of $B$ , note that $BP_0 \perp AC$ since $BCP_0 A$ is a kite. Thus, $BP_0$ bissects $AC$ , so $BP_0 = \tfrac{a\sqrt{2}}{2} + \tfrac{a\sqrt{6}}{2}$ . Additionally, $\angle BP_0 C = 45^\circ$ because $\angle BP_0 C$ bissects $\angle AP_0 C$ . Thus, the coordinates of $B$ are $(\tfrac{a+a\sqrt{3}}{2},\tfrac{a+a\sqrt{3}}{2})$ .

$[asy] import graph; size(9.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-1.2,xmax=15.2,ymin=-1.2,ymax=15.2; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype("2 2"); real gx=2,gy=2; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); xaxis(xmin,xmax,defaultpen+black,Arrows(6),above=true); yaxis(ymin,ymax,defaultpen+black,Arrows(6),above=true); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); pair A=(0,10),C=(10,0),P=(0,0),B=(13.660,13.660); dot(A); label("$A$",A,W); dot(C); label("$C$",C,S); dot(P); label("$P_0$",P,SW); dot(B); label("$B$",B,NE); draw(A--C--P--A); draw(A--B--C); draw(B--P,dotted); [/asy]$ By repeatedly applying the Midpoint Formula, we can determine the coordinates of $P_1$ , $P_2$ , $P_3$ , and so on. We can also use the Distance Formula to calculate the distance of $P_0P_1$ , $P_0P_2$ , and so on. The values are shown in the below table.

$n$	Coordinates of $P_n$	$P_0 P_n$
1	$(0,2a)$	$2a$
2	$(a+a\sqrt{3},-a+a\sqrt{3})$	$2a\sqrt{2}$
3	$(a-a\sqrt{3},a-a\sqrt{3})$	$a\sqrt{6} - a\sqrt{2}$
4	$(-a+a\sqrt{3},a+a\sqrt{3})$	$2a\sqrt{2}$
5	$(2a,0)$	$2a$
6	$(0,0)$	$0$
7	$(0,2a)$	$2a$
8	$(a+a\sqrt{3},-a+a\sqrt{3})$	$2a\sqrt{2}$

Note that the coordinates of $P_n$ as well as the distance $P_0 P_n$ cycle after $n = 6$ . Thus, $P_0P_n = \frac{\sqrt{6} - \sqrt{2}}{2} \cdot P_0P_1$ if $n \equiv 3 \pmod{6}$ , $P_0P_n = 0$ if $n \equiv 0 \pmod{6}$ , $P_0P_n = P_0P_1$ if $n \equiv 1, 5 \pmod{6}$ , and $P_0P_n = \sqrt{2} \cdot P_0P_1$ if $n \equiv 2, 4 \pmod{6}$ .

@@ Line 75: / Line 75: @@
 {{Pan African MO box
 |year=2001
-|before=First Problem
+|num-b=2
-|num-a=2
+|num-a=4
 }}
 [[Category:Olympiad Geometry Problems]]

2001 Pan African MO (Problems)
Preceded by Problem 2	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 4
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2001 Pan African MO Problems/Problem 3: Difference between revisions

Revision as of 00:15, 21 December 2019

Problem

Solution

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