CMU Putnam Seminar, geometry lecture

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CMU Putnam Seminar, Fall 2022
13 geometry lecture

• Let $P$ be a point inside a continuous closed curve in the plane which does not intersect itself. Show that there are two points on the curve whose midpoint is $P$.
  设 $P$ 为平面中不自交的连续闭合曲线内的点。证明曲线上有两个点的中点是$P$。
• Let convex quadrilateral $A B C D$ be given in a plane, and let $X$ be a point not on the plane. Show that there are points $A', B', C'$, and $D'$ on the lines $X A, X B, X C$, and $X D$, respectively, with the property that $A' B' C' D'$ is a parallelogram.
  设凸四边形$A B C D$在平面内，令$X$为不在平面内的点。证明在直线$X A、X B、X C$和$X D$上分别有点$A'、B'、C'$和$D'$，使 $A' B' C' D'$ 是平行四边形。
• Given any bounded plane region, prove that there are three concurrent lines that cut it into six pieces of equal area.
  给定任何有界平面区域，证明存在 3 条共点的直线将其切割成面积相等的 6 块。
• Given any bounded plane region, prove that there is a point through which no line trisects the area.
  给定任何有界平面区域，证明存在一点，通过该点的任何直线不会三等分面积。
• Given a finite collection of closed squares of total area 3, prove that they can be arranged to cover the unit square.
  给定总面积为 3 的有限个闭正方形，证明它们可以覆盖单位正方形。
• Given a finite collection of closed squares of total area $\frac{1}{2}$, prove that they can be arranged to fit in the unit square (with no overlaps).
  给定总面积为 $\frac{1}{2}$ 的有限个闭正方形，证明它们可以放入单位正方形而不重叠。
• Let $O A$ and $O B$ be two rays in the plane, and let $P$ be a point between them. Which point $X$ on the ray $O A$ has the property that if $X P$ is extended to meet the ray $O B$ at $Y$, then $X P \cdot P Y$ is minimized?
  设$O A$ 和$O B$ 是平面中的两条射线，$P$ 是它们之间的一个点。求射线 $O A$ 上的点 $X$，延长 $X P$ 与射线 $O B$ 相交于 $Y$，使 $X P \cdot P Y$ 最小？
• Given a region whose boundary is a simple polygon of area $a$ and perimeter $p$, prove that it contains a disc with radius larger than $a / p$.
  给定一个区域，其边界是面积 $a$ 和周长 $p$ 的简单多边形，证明它包含一个半径大于 $a / p$ 的圆盘。
• Given a right triangle and a finite set of points inside it, prove that these points can be connected by a path of line segments, such that the sum of squares of segment lengths in this path is at most the square of the hypotenuse.
  给定一个直角三角形和其中的有限个点，证明这些点可以通过一条线段路径连接起来，使得这些线段的长度的平方和至多为斜边的平方。
• Let an ellipse have center $O$ and foci $A$ and $B$. For a point $P$ on the ellipse, let $d$ be the distance from $O$ to the line of tangency to the ellipse at $P$. Show that $P A \cdot P B \cdot d^2$ is independent of the position of $P$.
  设一个椭圆的中心为 $O$，焦点为 $A$ 和 $B$。对于椭圆上的点 $P$，设 $d$ 为从 $O$ 到 $P$ 处的椭圆切线的距离。 证明 $P A \cdot P B \cdot d^2$ 与 $P$ 的位置无关。
  

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我们计算$P(x_0,y_0)$处的切线
$$
\frac{x_0}{a^2}x + \frac{y_0}{b^2}y = 1
$$
到焦点 $(c,0)$ 和 $(-c,0)$ 的距离$$
d_{1,2}=\frac{\frac{x_0c}{a^2}±1}{\sqrt{\frac{x_0^2}{a^4} + \frac{y_0^2}{b^4}}}
$$要证明它们的乘积为定值, 使用$a^2-b^2=c^2,$
\begin{align*}
d_1d_2=&\frac{1-\frac{c^2x_0^2}{a^4}}{\frac{x_0^2}{a^4}+\frac{y_0^2}{b^4}}\\
=&\frac{1-\frac{c^2x_0^2}{a^4}}{\frac{x_0^2}{a^4}+\frac1{b^2}\left(1-\frac{x_0^2}{a^2}\right)}\\
=&\frac{1-\frac{c^2x_0^2}{a^4}}{\frac1{b^2}\left(1-(a^2-b^2)\frac{x_0^2}{a^4}\right)}\\
=&b^2
\end{align*}
由\begin{array}l
d=\frac{d_1+d_2}2,\\
d_1d_2=b^2,\\
PA+PB=2a,\\
\frac{PA}{d_1}=\frac{PB}{d_2}\quad\text{(由光学性质)}
\end{array}得到\begin{align*}&P A \cdot P B \cdot d^2-a^2b^2\\=&\frac{1}{4} PA\cdot PB\left(d_1+d_2\over2\right)^2-\left(PA+PB\over2\right)^2 d_1d_2\\=&\frac14 \left(-d_2 PA + d_1 PB\right) \left(d_1 PA - d_2 PB\right)
\\=&0\end{align*}

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A Problem Seminar, D. J. Newman
57.（A CHEERFUL FACT ABOUT ...）Given a right triangle and a finite set of points inside it, prove that these points can be connected by a path of line segments the sum of whose squares is bounded by the square of the hypotenuse.
If, for example, the set consisted of the three vertices we would
choose the path made of the two legs and then the sum of the
squares would equal the square of the hypotenuse, but note that
the other path, one leg and the hypotenuse, would fail. So it is a
challenge; some care must be taken.
Aiming for an induction argument we drop the altitude from
the right angle and thereby break the original triangle into two
right triangles. But now, as is often the case in induction
arguments, we must prove a slightly harder result. Namely, let
us require also that our path begins at one end of the
hypotenuse and ends at the other. Say we so connect up the points in
our two triangles. Note that just joining these two paths will not
quite serve for the desired path in the original triangle. There is
now an extra point included, namely, the right angle vertex. But
we can get rid of it! Namely, consider the two segments that are
connected to this vertex, remove them, and insert the single
segment between the two points thus exposed. We have
removed two sides of a triangle with an acute angle between them
and replaced them by the third side. The sum of the squares has
gone down!
The induction is essentially accomplished except for the
nuisance that the number of points may not be sufficiently
diminished from the original triangle to the two new ones. This
situation is corrected, however, by continuing the construction
perhaps a few more times.

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A Problem Seminar, D. J. Newman
35. Given any bounded plane region, prove that there are three concurrent lines that cut it into six pieces of equal area.
We use a continuity argument, albeit a rather elaborate one. So pick a direction, $\theta$, and (by a continuity argument) produce the unique directed line in that direction which bisects the region. Next, through any point on this line we draw another directed line so that the forward wedge contains one-sixth of the region.

When this point is chosen all the way back on the line, then the backward wedge contains almost 0 of the region

When the point is all the way forward, this backward wedge contains almost one-half of the region.

Thus, by continuity we may choose to make the forward wedge and the backward wedge each contain one-sixth of the region.

This is our second line, and for the third line we can, in the same way, arrange for the forward and backward wedges to each contain one-third of the region.

These three lines are not necessarily concurrent (if they were the problem would be solved), but we define $f(\theta)$ as the directed length from the first intersection point (meeting point of the first line and the second line) to the second intersection point (meeting point of the first line and the third line).

Now when $\theta$ is replaced by $\theta+\pi$, the entire picture remains the same except for the orientation and order of the lines. In short, we obtain $f(\theta+\pi)=-f(\theta)$. Our final continuity argument say that $f(\theta)$, therefore, has a zero, and as we said before, this gives us our desired concurrent lines.

Borsuk–Ulam theorem #With odd functions

36. Given any bounded plane region, prove that there is a point through which no line trisects the area.
Draw the three lines à la Problem 35 . The point where they meet obviously has this property since any line through it has two wedges + some more on each side of it.

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A Problem Seminar, D. J. Newman
33. Given a region whose boundary is a simple polygon, prove that it contains a disc with radius larger than area/perimeter.
This problem can really be perplexing if one takes the wrong point of view. This wrong approach being the attempt to
construct this large disc. The passive (female?) approach is the one that succeeds here. Namely, we assume that there is no disc of radius $r$ lying in the region. Then every point in the region is within $r$ of the boundary. This means that if we take strips of width $r$ along each side they (essentially) cover the region.

Some trouble is encountered at the vertices, so we need to do some careful bookkeeping here. Namely, if we have a side from $A$ to $B$, then we erect not a strip, but the set within $r$ of $A B$ and also within the angle bisectors at $A$ and $B$. The area of this set is equal to$$r \times\text{length of }A B+\text{two contributions from the endpoints}$$

If the angle at the endpoint, say $d$, is $<\pi$ then we must subtract the right triangle of height $r$ and angle $A / 2$, and this has area bigger than the circular sector it contains, viz., $\left(r^2 / 4\right)(\pi-A)$. If the angle at the endpoint, say $B$, is $\geqslant \pi$, then we must add on the sector of angle $(B-\pi) / 2$, an area of $\left(r^2 / 4\right)(B-\pi)$. In all; then, the total area of these sets is bounded by$$r\times\text{perimeter}-\left(r^2 / 2\right) \sum\text{exterior angles}\coloneqq r \times \text{perimeter}-\pi r^2$$and since the union of these sets contains the whole region by hypothesis, we see that$$\text{Area of polygon} \leqslant r\times\text{perimeter}-\pi r^2<r\times\text{perimeter}$$This being true for all $r$ in excess of the inscribable radius shows that area/perimeter is a (strict) lower bound for this radius.

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A Problem Seminar, D. J. Newman
26. Given a finite collection of closed squares of total area 3, prove that they can be arranged to cover the unit square.
The construction we give will do this covering without "tilting" the squares and then it becomes a very pretty result because the 3 is best possible. Namely, 3 squares each of area $1-\varepsilon$ cannot cover: they cannot even cover the 4 vertices, since each covers at most one!

Sometimes the "dumbest" approach to a problem is the one that leads to success, and this happens to be the case here. Just arrange the squares in decreasing (non-increasing) order and line them up in rows until the length of each row first exceeds 1. We obtain, thereby, rows of length $>1$ and minimum heights which we may call $h_1, h_2, h_3, \ldots$ and finally a row of length $\leqslant 1$ with a height of perhaps 0 . We need only prove that $h_1+h_2+h_3+\cdots \geqslant 1$ for then the juxtaposition of these rows will cover the whole unit square. Now the maximum heights of these row's are bounded by $1, h_1, h_2$, etc., and so the areas of said rows are bounded by $1 \cdot\left(1+h_1\right), h_1\left(1+h_2\right), h_2\left(1+h_3\right), \ldots$ which are then bounded by $1+h_1, h_1+h_2, h_2+h_3, \ldots$. Comparing areas then gives
\[
1+2 h_1+2 h_2+\cdots \geqslant \text { Total Area } \geqslant 3
\]
or $h_1+h_2+\cdots \geqslant 1$, as desired.

27. Given a finite collection of squares of total area $\frac12$, show that they can he arranged so as to fit in a unit square (with no overlaps).
Since two squares of side $\frac{1}{2}+\varepsilon$ cannot be so fit into the unit square, this is also a best possible result.

We proceed as in Problem 26, arranging the squares in decreasing order and then making rows of them this time, however, we stop each row just before its length exceeds 1. So, calling the maximum heights $H_1, H_2, \ldots$, we see that this time we want the inequality $H_1+H_2+\cdots \leqslant 1$ (so that the "pile up" of all the rows will then fit into the unit square).

To obtain this inequality, we call the lengths of the rows $L_1, L_2, \ldots$, respectively, and note that each $L_k+H_{k+1}>1$. Hence, $L_k+H_{k+1}-H_k>1-H_k \geqslant 1-H_1$, and so $H_{k+1}\left(L_k+H_{k+1}-H_k\right)>\left(1-H_1\right) H_{k+1}$. If we sum these inequalities we note that the left-hand sides add up to an area bounded by the full sum of the individual squares less the first one; i.e., to $\frac{1}{2}-H_1^2$. So we have $\frac{1}{2}-H_1^2 \geqslant\left(1 -H_1\right)\left(H_2+H_3+\cdots\right)$ and this, in turn, gives $\frac{1}{2}+H_1-2 H_1^2 \geqslant\left(1-H_1\right)\left(H_1+H_2+H_3+\cdots\right)$. The proof is completed by noting that $\frac{1}{2}+H_1-2 H_1^2=1-H_1-\frac{1}{2}\left(1-2 H_1\right)^2 \leqslant 1-H_1 ;$ which gives our required inequality, namely, $1-H_1 \geqslant\left(1-H_1\right)\left(H_1+H_2+H_3+\cdots\right)\implies1\geqslant H_1+H_2+H_3+\cdots$.

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Show we can find two points on the curve whose midpoint is P.
Let $\mathrm{C}$ be the curve and $\mathrm{C}^{\prime}$ the curve obtained by rotating $\mathrm{C}$ through $180^{\circ}$ about $\mathrm{P}$. Let $\mathrm{m}$ be a point on $\mathrm{C}$ closest to $\mathrm{P}$, and $\mathrm{M}$ a point on $\mathrm{C}$ furthest from $\mathrm{P}$. Then $\mathrm{m}$ must lie inside or on $\mathrm{C}^{\prime}$, and $\mathrm{M}$ must lie outside or on $\mathrm{C}^{\prime}$. Hence $\mathrm{C}$ and $\mathrm{C}^{\prime}$ must intersect. Take $\mathrm{Q}$ to be any common point. Then the point $\mathrm{Q}^{\prime}$ obtained by rotating $\mathrm{Q}$ through $180^{\circ}$ must also lie on $\mathrm{C}$ and $\mathrm{C}^{\prime}$. Now $\mathrm{P}$ is the midpoint of $\rm QQ'$.

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$\mathbb{R}^3$中4个点必线性相关, 所以存在不全为零的$a, b, c, d \in
\mathbb{R}$使$a \overrightarrow{XA} + b \overrightarrow{XB} + c
\overrightarrow{XC} + d \overrightarrow{XD} = 0$, 即$a
\overrightarrow{XA} - (- b) \overrightarrow{XB} = (- c)
\overrightarrow{XC} - d \overrightarrow{XD}$.将$A, B, C,
D$关于$X$位似$a, - b, - c, d$倍到$A', B', C', D'$, 则$A' B' C'D'$是平行四边形.

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设$\angle AOP=\alpha,\angle POB=\beta,\angle OPX=x$,由正弦定理, $\frac{PX}{PO} = \frac{\sin \alpha}{\sin (x +
\alpha)}, \frac{PY}{PO} = \frac{\sin \beta}{\sin (x - \beta)}$,
求$\frac{\sin \alpha}{\sin (x + \alpha)}  \frac{\sin \beta}{\sin (x -
\beta)}$最小值, 就是求$\sin (x + \alpha) \sin (x - \beta),\beta < x < \pi - \alpha$的最大值.
$2\sin (x + \alpha) \sin (x - \beta)=\cos(\alpha-\beta)-\cos(2x+\alpha-\beta)\le\cos(\alpha-\beta)+1$
$$\cos(2x+\alpha-\beta)=1⇔2x+\alpha-\beta=π⇔x=\frac{π-\alpha+\beta}2⇔OX=OY$$

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hbghlyj 发表于 2022-12-16 13:34
If we sum these inequalities we note that the left-hand sides add up to an area bounded by the full sum of the individual squares less the first one

设第$k$行的第一个正方形是第$n_k$个正方形(则$n_1=1$),第$n$个正方形的边长是$a_n$.
因为$L_k$是第$k$行的正方形的边长之和,$L_k=a_{n_k}+\dots+a_{n_{k+1}-1}$.
因为$H_k$是第$k$行的第一个正方形,$H_k=a_{n_k}$.
因为$a_{n_k+1}\ge\dots\ge a_{n_{k+1}}$和
$$L_k+H_{k+1}-H_k=(a_{n_k}+\dots+a_{n_{k+1}-1})+a_{n_{k+1}}-a_{n_k}=a_{n_k+1}+\dots+a_{n_{k+1}}$$所以
$$H_{k+1}(L_k+H_{k+1}-H_k)=a_{n_{k+1}}(a_{n_k+1}+\dots+a_{n_{k+1}})\le a_{n_k+1}^2+\dots+a_{n_{k+1}}^2$$Summing over $k\ge1$, we get
$$\sum_{k\ge1}H_{k+1}(L_k+H_{k+1}-H_k)\le\sum_{n\ge2} a_n^2$$