Trigonometry

Using trigonometry we can derive that $CH = 2R \cos C$ so $CH^2 = 4R^2 \cos ^2 C$ . Now $A+B+C = 180^{\circ}$ so:

$\tan A + \tan B + \tan C = \tan A\tan B\tan C$

then substituting in and solving gives $\tan C = 3$ . Therefore $\sec ^2 C = \tan ^2 C + 1 = 10 \implies \cos ^2 C = \frac{1}{10}$ .

Therefore $CH^2 = 4 \times 5^2 \times \frac{1}{10} = \boxed{10}$ .

suppose O is the circumcenter.then <BOC=2<BAC.now tan<BAC=1 implies <A=45.Let extended CH cuts AB at C' and BH cuts AC at B'.Now HC'=BC'(=x) and HB'=CB'(=y).Now tan<B=2 implies (\sqrt { 2 } y+x)/x=2,or \sqrt { 2 } y=x.It's a well known fact that AH=2.OS(S is the feoot of the perpendicular from O to BC.But \sqrt { 2 }.OS=5.Using pythagoras theorem on triangle AC'H we get (x+\sqrt { 2 }.y)^2+x^2=50 we get 2.y^2=10

tanA = 1, So, A = 45

tanB = 2, So, B = atan(2) = 63.434

C = 180 - (A+B) = 180-108.434 = 71.565

CH = 2R cos C

= 2 * 5 * cos(71.565)

= 10*0.3162

= 3.1622

CH^2 = 3.1622^2 = 10.

from tan(A)=1 and tan(B)=2, we know that: sin(A)=1/sqrt(2), sin(B)=2/sqrt(5), sin(C)=3/sqrt(10)

a/sin(A)=b/sin(B)=c/sin(C)=2R=10, with a, b and c the lengths of the sides of the triangle, this gives: a=5 sqrt(2), b=4 sqrt(5), c=3 sqrt(10)

Define a catresian system so that (keep in mind that the angle in A is pi/4 or 45°): A=(0, 0, 0), B=(3 sqrt(10), 0), C=(2 sqrt(10),2 sqrt(10))

orthocentre through A is given by y=x/2

orthocentre through C is given by x=2 sqrt(10)

Coördinates of H, on the intersection of these two lines, are therefore (2 sqrt(10), sqrt(10))

d(C,H)=2 sqrt(10) - sqrt(10) = sqrt(10) or CH² = 10

$\text{D and F are the feet of the altitudes of triangle ABC as shown on the sketch.}\\ \text{Let AB=3k where k is the scale factor. Since } ~~~ ~A=Tan^{-1}1 ~~~ ~ and ~ ~~~ B=Tan^{-1}2,\\ \therefore ~AF=2k, ~~~~~~~~FB=k, ~~~~~~~~ CF=2k.\\ Applying ~ Pythagoras ~ to ~ \Delta s ~ CAF ~ and ~CFB, ~~~~~~ CA=2*\sqrt2*k,~~~~~ ~CB=\sqrt5k.\\ Applying ~Extended ~ Sin ~ Law~to ~\Delta ~ABC,~~~~~~ 2*R=10=\dfrac{CB}{SinA}=\sqrt5k*\sqrt2.\\ \implies ~\color{#3D99F6}{k=\sqrt{10} }.\\ In~ the ~rt.\angle ed ~ \Delta s ~ CAD ~ and ~CHD, ~~~~~~~~CA*CosC=CD=HC*SinB,\\ \therefore ~HC=\dfrac{CA*Cos(180-A-B)}{SinB}= \dfrac{2*\sqrt2*\sqrt{10}*Cos(180-A-B)}{SinB}.\\ \implies HC^2=\huge \color{#D61F06}{10}$

Note:- angle A=arcTan1=45, angle B=arcTan2=63.4349.

I think everybody is using own way for this question. Therefore, I should also tell how I did this.

Start from the end, let me write about the last steps first. Assign A as origin.

We shall obtain A (0, 0), B ( $3 \sqrt{10}$ , 0) and C (2 $\sqrt{10}$ , 2 $\sqrt{10}$ ). Solve by geometry:

$\frac{y - 0}{x - 3 \sqrt{10}} = \frac{2 \sqrt{10}- 0}{2 \sqrt{10} - 3 \sqrt{10}} = \frac{2 \sqrt{10}}{-\sqrt{10}}$ = -2 is the line B C.

Perpendicular line to B C from A (0, 0) crossing it should have a gradient of $\frac12$ for $m_1 m_2$ = -1.

Therefore, this line y = $\frac12$ x is intercepting with the line of C (2 $\sqrt{10}$ , 2 $\sqrt{10}$ ) to the x-axis.

At x = 2 $\sqrt{10}$ , y = $\sqrt{10}$ . This is point H.

Obviously, for C H parallel to the y-axis, C H = 2 $\sqrt{10}$ - $\sqrt{10}$ = $\sqrt{10}$ .

Therefore $(CH)^2$ = $(\sqrt{10})^2$ = 10.

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$Resources:$

Given that tan A = 1 and tan B = 2,

tan C = tan (180 $^\circ$ - A - B) = - tan (A + B) = - $\frac{tan A + tan B}{1 - tan A tan B}$ = 3.

$sin 2 x$ = $\frac{2 tan x}{1 + tan^2 x}$

sin 2 A = 1 and sin A = $\frac{1}{\sqrt2}$

sin 2 B = $\frac45$ and sin B = $\frac{2}{\sqrt5}$

sin 2 C = $\frac35$ and sin C = $\frac{3}{\sqrt{10}}$

a b $\frac{3}{\sqrt{10}}$ = b c $\frac{1}{\sqrt{2}}$ = a c $\frac{2}{\sqrt{5}}$ = $a h_a = b h_b = c h_c$ = $5^2$ (1 + $\frac45$ + $\frac35$ ) = 60

Therefore,

a = $5 \sqrt2$ and $h_a = 6 \sqrt2$

b = $4 \sqrt5$ and $h_b = 3 \sqrt5$

c = $3 \sqrt{10}$ and $h_c = 2 \sqrt{10}$

$\angle$ A = 45 $^\circ$ makes coordinates of C an ease to determine.

b = AC = $4 \sqrt5$

$\implies A C' = \frac{AC}{\sqrt2} = 2 \sqrt{10}$ ,

Therefore, C is located at ( $2 \sqrt{10}$ , $2 \sqrt{10}$ ).

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This solution takes $no$ special theorem as resources for answering the question.

Answer: $\boxed{10}$

Since other solutions have been taken, we will use barycentric coordinates.

From the relation $\tan A + \tan B + \tan C = \tan A \tan B \tan C$ we find $\tan C = 3$ .

Thus, $H = (\tan A: \tan B: \tan C) = (1 : 2 : 3) = (\frac{1}{6} , \frac{1}{3} , \frac{1}{2})$

This implies that $\begin{aligned} \vec{CH} & = (\frac{1}{6} , \frac{1}{3}, - \frac{1}{2}) \\ |\vec{CH}|^2 & = -a^2yz - b^2xz - c^2xy = \frac{a^2}{6} + \frac{b^2}{12} - \frac{c^2}{18} \end{aligned}$

However, will $\tan$ , we can derive $\sin$ and the side lengths easily. $\begin{aligned} a & = 2R\sin A = 5\sqrt{2} \\ b & = 2R\sin B = 4\sqrt{5} \\ c & = 2R\sin C = 3 \sqrt{10} \end{aligned}$ Substituting these values, we get that $CH^2 = \boxed{10}$ .