====== Plane Geometry Problems ======

===== Q.2 =====
[[https://www.geogebra.org/m/edrbbhkv|如图]], $AB=BC=3, CD=4, DE=5, BC \perp CD, CD\perp DE, AB \perp AE, 求AE 的长度。$

&GGB &PP
++++ Solution 1|
过 $B$ 作 $CD$ 的平行线交 $ED$ 延长线与 $F$, 链接 $BE$, 则 $BCDF$ 为矩形。

在直角 $\triangle BFE$ 中，$BF=CD=4, FE=FD+DE=BC+ED=3+5=8$

$\therefore BE^2=BF^2+FE^2=16+64=80$

在直角 $\triangle ABE$ 中，$AE^2=BE^2-AB^2=80-9=71$

$\therefore AE=\sqrt{71}$
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&GGB &PP
++++ Solution 2|
以 $B$ 为原点，$BC$ 为 $y$ 轴，建立坐标系，令 $\angle ABC$ 为 $\theta$,则有各点坐标：

$A(-3\ast\sin(\theta), 3\ast\cos(\theta)), B(0,0), C(0,3), D(4,3), E(4,8)$

$\because AE\perp AB$, 两线斜率乘积为 $-1$

$\therefore \frac{8-3\ast\cos(\theta)}{4+3\ast\sin(\theta)}\ast\frac{3\ast\cos(\theta)}{-3\ast\sin(\theta)}=-1$

$\therefore 24\cos(\theta)-9\cos^2(\theta)=12\sin(\theta)+9\sin^2(\theta)$

$\therefore 8\cos(\theta)-4\sin(\theta)=3$

$AE^2=(8-3\cos(\theta))^2 + (4+3\sin(\theta))^2 =89-6(8\cos(\theta)-4\sin(\theta)) = 89-18=71$

$\therefore AE=\sqrt{71}$
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&MMA

++++ Show/Hide|
<code mathematica>
scene = GeometricScene[{{a, b, c, d, e}, {x}},
                       {EuclideanDistance[a, b] == EuclideanDistance[b, c] == 3,
                        EuclideanDistance[c, d] == 4, EuclideanDistance[d, e] == 5,
                        EuclideanDistance[e, a] == x,
                        PolygonAngle[{a, b, c, d, e}, a] == 90 Degree,
                        PolygonAngle[{a, b, c, d, e}, c] == 270 Degree,
                        PolygonAngle[{a, b, c, d, e}, d] == 90 Degree}];
(* RandomInstance[scene] *)
GeometricSolveValues[scene, x]
(* {Sqrt[71]} *)
</code>
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----

===== Q.1 =====
$P$ is a point inside $\square ABCD$. The distances from $P$ to three vertices are 1, 2, and 3, respectively. Find the distance from $P$ to the fourth vertex, and the side length of the square.

&GGB &PP
++++ Show/Hide|

(1)

[[https://www.geogebra.org/m/fy2yr4h7|如图]]，过 $P$ 做 $MN // AB$ 交 $AD$ 于 $M$，交 $BC$ 于 $N$，则有：$AM=BN, DM=CN$

$AP^2 - AM^2 = DP^2 - DM^2, BP^2 - BN^2 = CP^2 - CN^2$

$1^2 - AM^2 = DP^2 - DM^2, 2^2 - AM^2 = 3^2 - DM^2$

两式相减得 $DP^2=6, DP=\sqrt{6}$

(2)

[[https://www.geogebra.org/m/fy2yr4h7|如图]]，
$AP=1, BP=2, CP=3,$
过 $B$ 做 $BE \perp BP$ 且取 $BE=BP$，连接 $AE, PE$.

$\because \angle ABP + \angle CBP = \angle ABP + \angle ABE = 90^{\circ}$

$\therefore \angle CBP = \angle ABE $

又 $\because CB=AB, PB=EB$

$\therefore \triangle ABE \cong \triangle CBP$

$\therefore AE=CP=3$

在等腰直角 $\triangle BPE$ 中，$PE=\sqrt{2}PB=2\sqrt{2}$

$\therefore AP^2 + PE^2 = 1 + 8 = AE^2$

$\therefore \angle APE = 90^{\circ}, \angle APB = 90^{\circ} + 45^{\circ} = 135^{\circ}$

$\therefore AB = \sqrt{AP^2 + PB^2 - 2\ast AP\ast BP\ast\cos(135^{\circ})} = \sqrt{5+2\sqrt{2}}$

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&MMA
++++ Show/Hide|
<code mathematica>
g = GeometricScene[
      {{a, b, c, d, p}, {x, y}},
      {PolygonAngle[{a, b, c, d}, a] == 90 Degree, 
       PolygonAngle[{a, b, c, d}, b] == 90 Degree, 
       PolygonAngle[{a, b, c, d}, c] == 90 Degree, 
       EuclideanDistance[a, b] == EuclideanDistance[b, c],
       EuclideanDistance[p, a] == 1, EuclideanDistance[p, b] == 2, EuclideanDistance[p, c] == 3,
       EuclideanDistance[p, d] < Sqrt[2] y, (* p is inside the square*)
       x == EuclideanDistance[p, d], y == EuclideanDistance[a, b]},
      {}
    ]
GeometricSolveValues[g, {x, y}]
(* we get: {{Sqrt[6], Sqrt[5 + 2 Sqrt[2]]}} *)
</code>

Furthermore, you can use ''RandomInstance[g]'' to get a graph of the scene.
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