Pythagorean Theorem: Very Difficult Problems with Solutions

Solution:
Let [tex]p[/tex], [tex]q[/tex], [tex]r[/tex], [tex]s[/tex] be the distances from [tex]E[/tex] to the sides [tex]AB[/tex], [tex]CD[/tex], [tex]AD[/tex], [tex]BC[/tex]. Drawing the perpendiculars from [tex]E[/tex] to the sides, we get four right triangles, one at each vertex. By the Pythagorean theorem:
[tex]AE^2=p^2+r^2[/tex], [tex]CE^2=q^2+s^2[/tex]
[tex]DE^2=q^2+r^2[/tex], [tex]BE^2=p^2+s^2[/tex]
Therefore [tex]AE^2+CE^2=p^2+q^2+r^2+s^2=DE^2+BE^2=9+49=58[/tex]

Solution:
Let [tex]G[/tex] be the intersection point of the medians. It divides each median in the ratio [tex]2:1[/tex] from the vertex. Let [tex]GN=x[/tex] and [tex]GK=y[/tex]; then [tex]AG=2x[/tex], [tex]BG=2y[/tex], and the angle at [tex]G[/tex] is right. Also [tex]BN=3[/tex] and [tex]AK=4[/tex].
[tex]\triangle BGN:\ (2y)^2+x^2=3^2=9[/tex]
[tex]\triangle AGK:\ (2x)^2+y^2=4^2=16[/tex]
[tex]5x^2+5y^2=25 \Rightarrow x^2+y^2=5[/tex]
[tex]AB^2=AG^2+BG^2=4x^2+4y^2=20 \Rightarrow AB=\sqrt{20}=2\sqrt{5}[/tex]

Solution:
Let [tex]AC=6[/tex], [tex]BC=8[/tex], [tex]CH[/tex] the altitude and [tex]CM[/tex] the median ([tex]H[/tex] and [tex]M[/tex] on [tex]AB[/tex]).
[tex]AB^2=6^2+8^2=100 \Rightarrow AB=10[/tex], [tex]AM=5[/tex]
[tex]\dfrac{6\times 8}{2}=\dfrac{10\times CH}{2} \Rightarrow CH=\dfrac{24}{5}[/tex]
[tex]AH^2=AC^2-CH^2=36-\dfrac{576}{25}=\dfrac{324}{25} \Rightarrow AH=\dfrac{18}{5}[/tex]
[tex]HM=AM-AH=5-\dfrac{18}{5}=\dfrac{7}{5}[/tex]
[tex]S_{CHM}=\dfrac{1}{2}\times \dfrac{7}{5}\times \dfrac{24}{5}=\dfrac{84}{25}[/tex]

Solution:
Let the segments be [tex]x[/tex] and [tex]x+7[/tex]; then the altitude is [tex]h=x+3[/tex]. By the Pythagorean theorem in the two small triangles and in the big one:
[tex](h^2+x^2)+\bigl(h^2+(x+7)^2\bigr)=(2x+7)^2[/tex]
[tex]2h^2+2x^2+14x+49=4x^2+28x+49 \Rightarrow h^2=x^2+7x=x(x+7)[/tex]
[tex](x+3)^2=x^2+7x \Rightarrow 6x+9=7x \Rightarrow x=9[/tex]
So [tex]h=12[/tex], the segments are [tex]9[/tex] and [tex]16[/tex], and the legs are
[tex]\sqrt{12^2+9^2}=15[/tex], [tex]\sqrt{12^2+16^2}=20[/tex]

Solution:
The median to the hypotenuse equals half of the hypotenuse, so the two triangles have sides [tex]a, \frac{c}{2}, \frac{c}{2}[/tex] and [tex]b, \frac{c}{2}, \frac{c}{2}[/tex]:
[tex]a+c=8[/tex], [tex]b+c=9[/tex]
By the Pythagorean theorem: [tex](8-c)^2+(9-c)^2=c^2[/tex]
[tex]64-16c+c^2+81-18c+c^2=c^2 \Rightarrow c^2-34c+145=0 \Rightarrow (c-5)(c-29)=0[/tex]
For [tex]c=29[/tex] the legs would be negative, so [tex]c=5[/tex], [tex]a=3[/tex], [tex]b=4[/tex].

Solution:
The median to the hypotenuse is [tex]\frac{c}{2}[/tex], so the altitude is [tex]h=\frac{c}{2}-7[/tex]. Expressing the area in two ways, [tex]ab=ch[/tex]:
[tex]ab=c\left(\dfrac{c}{2}-7\right)=\dfrac{c^2}{2}-7c[/tex], [tex]a+b=72-c[/tex]
[tex](a+b)^2=a^2+b^2+2ab \Rightarrow (72-c)^2=c^2+c^2-14c[/tex]
[tex]5184-144c+c^2=2c^2-14c \Rightarrow c^2+130c-5184=0[/tex]
[tex](c-32)(c+162)=0 \Rightarrow c=32[/tex]