Isl 2003 c3

by Wolstenholme, Aug 5, 2014, 5:08 AM

Let $n \geq 5$ be an integer. Find the maximal integer $k$ such that there exists a polygon with $n$ vertices (convex or not, but not self-intersecting!) having $k$ internal $90^{\circ}$ angles.

Solution:

I shall show that when $n = 5$ we have $k = 3$ and when $n > 5$ we have $k = \left\lceil\frac{2n + 1}{3}\right\rceil$ .

Lemma 1: When $n = 5$ we have $k = 3$

Proof: First I shall construct an example where the number of interior right angles is $3$ . Consider a square $ABCD$ . Let $E$ be a point not in the interior of $ABCD$ such that $AED$ is an isosceles right triangle with hypotenuse $AD.$ Then the pentagon $ABCDE$ clearly has $3$ interior right angles as desired. Now, I shall show that there may not be $4$ interior right angles. Assume the contrary. Then since the sum of interior angles must be equal to $540$ , we have that the non-right interior angle in the pentagon has angle measure of 180 degrees, contradiction. Similarly $5$ interior right angles are impossible so in this case $k = 3$ as desired.

Lemma 2: For all integers $n \geq 5$ , we have that $k \leq \left\lceil\frac{2n + 1}{3}\right\rceil$ .

Proof: Consider a $n$ -gon with $k$ interior right angles. Then the sum of the non-right interior angles is $180(n - 2) - 90k$ . But every such angle has angle measure less than 360 degrees so we have the inequality $180(n - 2) - 90k < 360(n - k)$ which yields the desired result.

Lemma 3: When $n = 6$ we can construct a figure with $5$ interior right angles.

Proof: Consider a square $ABCD$ . Construct points $E, F, G, H$ all distinct from $A, B, C, D$ such that quadrilaterals $ABEF$ and $BCGH$ are squares. Then the hexagon $BEFDGH$ has $5$ interior right angles as desired.

Lemma 4: When $n = 7$ we can construct a figure with $5$ interior right angles.

Proof: Consider hexagon $BEFDGH$ from the $n = 6$ case. Let $M, N$ be the midpoints of segments $BE, BH$ respectively. Then heptagon $DGHNMEF$ has $5$ interior right angles as desired.

Lemma 5: When $n = 8$ we can construct a figure with $6$ interior right angles.

Proof: Consider hexagon $BEFDGH$ from the $n = 6$ case. Let $M$ be the midpoint of segment $EF$ . Construct points $R, S$ not in the interior of the hexagon such that quadrilateral $FMRS$ is a square. Then octagon $DGHBEMRS$ has $6$ interior right angles as desired.

Lemma 6: For any $n$ -gon with $k$ interior right angles, we can construct an $(n + 3)$ -gon with $k + 2$ interior right angles.

Proof: Let the $n$ -gon be denoted $A_{1}A_{2} \dots A_{n}$ . Assume WLOG that $\angle{A_{1}A_{2}A_{3}} \ne 90$ . Let $X$ be a point distinct from $A_2$ on segment $A_{1}A_{2}$ and let $Y$ be a point distinct from $A_2$ on segment $A_{3}A_{2}$ . Construct points $V, W$ not in the interior of the $n$ -gon such that quadrilateral $XYVW$ is a square. Then the $(n + 3)$ -gon $A_{1}XWVYA_{3}A_{4} \dots A_n$ has $k + 2$ interior right angles as desired. By making the distances $A_{2}X$ and $A_{2}Y$ sufficiently small we can force the resulting $(n + 3)$ -gon to not self-intersect and so we have the desired result.

The combination of these lemmas clearly proves my initial claim. This problem is a good exercise in precision and rigor... not so much in skill

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glad to have found a fellow chipotle lover <3
by nukelauncher, Aug 13, 2020, 6:40 AM
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by fukano_2, Jun 14, 2020, 6:24 AM
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by songssari, Jun 12, 2016, 8:21 AM
shouts make blogs happier
by briantix, Mar 18, 2016, 9:57 PM
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by mishka1980, Sep 12, 2015, 10:33 PM
This is late, but where is the ARML results post?
by donot, Aug 31, 2015, 11:07 PM
"I am Sam"
"Sam I am"
by mathwizard888, Aug 12, 2015, 9:13 PM
HW $\textcolor{white}{}$
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
Are you Sam?
by Eugenis, Apr 4, 2015, 2:05 AM
@Brian: yes, yes I did #whoneedsalgskillz?

@gauss1181; hey!
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 2003 c3

by Wolstenholme, Aug 5, 2014, 5:08 AM

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