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I've been asked to help with the following school problem on geometry.

In the triangle $\Delta ABC$ one has $AB = 60$, $AC = 80$. Point $O$ is the centre of the circumscribed circle. Point $D$ belongs to the side $AC$. Additionally, one has $AO \perp BD$. One is asked to find $CD$.

(just in case, the answer is $35$)

I am really puzzled, since the information given clearly does not fix the triangle. I know how to solve the problem under the assumption that point $O$ belongs to $BD$. In this case, the solution goes as follows:

Denote $\alpha = \angle OAC$, $\beta = \angle OBC$.

  1. $\angle ACB = \dfrac{1}{2}\angle AOB = 45^\circ$.
  2. From the sum of angles of the triangle $\triangle ABC$, one has: $$\alpha + \beta = 45^\circ$$
  3. The law of sines for the triangle $\triangle ABC$ gives: $$\dfrac{AC}{\sin(\beta + 45^\circ)}=\dfrac{AB}{\sin(45^\circ)}$$ From where one can find $\beta$: $$\beta = \arccos\left( \dfrac{2\sqrt2}{3} \right) + 45^\circ$$
  4. From the triangle $\triangle AOD$ one finds: $$CD = AC - AD = AC - \dfrac{AO}{\cos(\alpha)}= AC - \dfrac{AO}{\cos(45^\circ - \beta)}$$

Substituting the value of $\beta$ indeed gives $CD = 35$.

Now, I have two questions:

  1. Is it possible to get the answer without the assumption I have made (or any other one).

  2. Can anyone present an easier solution? (just in case, this is one of $26$ problems in the $9$th grade quiz in Russian middle school — students are obviously limited in time and are not supposed to use Mathematica and even Stack Exchange)

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  • $\begingroup$ 1. Yes, it is possible to get the answer without the assumption you have made. 2. See my solution. $\endgroup$ – Futurologist Feb 24 '17 at 3:58
  • $\begingroup$ That's interesting. Either you or @kotomord has made a mistake then :) $\endgroup$ – mavzolej Feb 24 '17 at 4:58
  • $\begingroup$ my solution is the right one. $\endgroup$ – Futurologist Feb 24 '17 at 13:50
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Draw the line $AO$ and let $E$ be the second point of intersection of $AO$ with the circumcircle of triangle $ABC$ (the first point of intersection being $A$). Then $AE$ is the diameter of the circumcircle and therefore triangle $ABE$ is a right triangle ( $\angle \, ABE = 90^{\circ}$ ). Let $H$ be the itnersection point of $BD$ and $AO$.

enter image description here

Since by assumption $BD$ is orthogonal to $AO$, and therefore orthogonal to $AE$, segment $BD$ is an the altitude of $ABE$ through $B$. Hence triangles $AHB$ and $ABE$ are similar and thus $$\frac{AH}{AB} = \frac{AB}{AE}$$ which is equivalent to $$AH \cdot AE = AB^2 = 60^{2}$$ Triangle $AEC$ is right triangle ( $\angle \, ACE = 90^{\circ}$ ) and so is $AHD$, which means they are similar and thus $$\frac{AD}{AE}=\frac{AH}{AC} $$ which is equivalent to $$AD \cdot AC = AH \cdot AE = 60^2$$ and since $AD = AC - CD = 80 - CD$ and $AC = 80$ get the equation $$(80-CD)\cdot 80 = 60^2$$ When you solve it you get $CD = 35$.

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  • $\begingroup$ Did you assume here that the point $O$ belongs to $BD$? Also, "Let $H$ be the intersection point of $AD$ and $AO$" sounds a bit strange. Why isn't the intersection point just $A$? $\endgroup$ – mavzolej Feb 24 '17 at 4:50
  • $\begingroup$ @mavzolej Because it is a typo. It's $BD$ intersects $AO$ at $H$. I fixed it. Read it again. $\endgroup$ – Futurologist Feb 24 '17 at 13:47
  • $\begingroup$ I think you may also have a typo when saying "triangles $AHB$ and $ABC$ are similar". $\endgroup$ – mavzolej Feb 24 '17 at 16:40
  • $\begingroup$ @mavzolej I think it is obvious $\endgroup$ – Futurologist Feb 24 '17 at 17:39
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enter image description here

Let $G$ be the point of intersection of $AO$ and $BD.$

$AGB$ is a right triangle.

Drop altitudes $OE$ and $OF$ to sides $AB$ and $AC$ respectively

Since $AOB$ and $AOC$ are isosceles, these altitudes bisect their respective sides.

$AFO$ is similar to $AGB$

$AF:AO = AG: AB$

$AO\cdot AG = 1800$

$AEO$ is similar to $AGD$

$AE:AO = AG: AD$

$AE\cdot AD = 1800$

$AD = 45$

$CD = 35$

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  • $\begingroup$ Why is $AE = \dfrac{1}{2} AC$? The next line is also very strange. $\endgroup$ – mavzolej Feb 24 '17 at 1:47
  • $\begingroup$ @DougM AOB is an isosceles right triangle That's an isosceles triangle, indeed, but it's only right if $AO \cap BD \equiv O\,$. $\endgroup$ – dxiv Feb 24 '17 at 2:21
  • $\begingroup$ Also, last two lines clearly contain typos. $\endgroup$ – mavzolej Feb 24 '17 at 4:55
  • $\begingroup$ @dxiv thanks, this is cleaner. $\endgroup$ – Doug M Feb 24 '17 at 5:03
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    $\begingroup$ $AE = 40$ since it is the midpoint of $AC, AD\cdot AE = 1800, AD = 1800/80 = 45, CD = AC - AD = 80-45 = 35$ $\endgroup$ – Doug M Feb 24 '17 at 6:49
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I don't see how you can solve it without your assumption.

With regards to your method, I believe there is a marginally quicker means:

Note that $\triangle OAB$ is an isosceles right angled triangle $\Rightarrow \angle OAB = \angle ABO = \pi/4$

$AO = BO = 60cos(\pi/4) = 30\sqrt{2}$

Using your definition of $\alpha$, $AD cos\alpha = 30\sqrt{2}$

But then I can't find a way of avoiding a messy means of calculating $cos\alpha$

I think it's just a poor question in the context given.

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Imho, No.

In coordinates: Let $A(0, 0), C(80, 0), B(60 cos(f), 60sin(f) )$

Coordinates of $O$ is $(40, t)$. from $AO = BO$ $1600 + t^2 = 3600 - 120*sin(f)*t + t^2$ => $t = \frac{50}{3*sin(f)}$

Equation of $AO$ is $5x - 12*sin(f)*y = 0$, so, equation of BD is $12*sin(f)*x + 5*y = 60*12*sin(f)*cos(f) + 60*5*sin(f)$

So, coordinates of D is $60*cos(f) + 20$, and $CD = 60(1-cos(f))$

Maybe, I was create same typos and fails, but this method must work :)

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  • $\begingroup$ So you're saying that for a certain value of the angle $f$, your general solution turns into mine? But generally the answer is different? $\endgroup$ – mavzolej Feb 24 '17 at 1:58
  • $\begingroup$ Yes, if it is not my typos if solution $\endgroup$ – kotomord Feb 24 '17 at 2:03
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    $\begingroup$ @mavzolej Just to let you know, the answer above is wrong. The problem is correctly stated and the answer depends only on the lengths of segments $AB$ and $AC$. Basically if you vary the angle $A$ of the triangle but keep the lengths of $AB$ and $AC$ fixed, the length of the segment $CD$ is always the same. One does not need to assume anything about $ABO$! $\endgroup$ – Futurologist Feb 24 '17 at 3:57

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