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$\triangle ABC$ is isosceles with $AB = AC$ ans $\angle A = 20^{\circ}$ and $BC = 12$. $BP \perp AC$ ans $Q$ is a point on $AB$ such that $QB=6$. Find $\angle CPQ$.

I was not able to solve this problem. Tried it many times. It can be solved very easily if I use trigonometric functions and calculator. But I need to solve it in Olympiad Math's way. The Diagram is something like this -
enter image description here

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  • $\begingroup$ What is the way you need to solve this? $\endgroup$ Commented Dec 30, 2016 at 17:52
  • $\begingroup$ @Fimpellizieri Just without using Calculator. $\endgroup$ Commented Dec 30, 2016 at 17:52
  • $\begingroup$ the answer is $$120^{\circ}$$ $\endgroup$ Commented Dec 30, 2016 at 18:15
  • $\begingroup$ I found that same answer using laws of cosines but this is terribly long. $\endgroup$ Commented Dec 30, 2016 at 19:16
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    $\begingroup$ Let $X \in AP$ be such that $\angle CBX=20^{\circ}$. Then $\triangle CBX$ is isosceles and $BX=BC=12$. Now let $Y \in AB$ be such that $\angle BYX=90^{\circ}$. We claim that $Y=Q$. Indeed, $\angle BXA=110^{\circ}$, so $\angle AXY=80^{\circ}$ and hence $\angle BXY = 30^{\circ}$. It follows that $\triangle BXY$ is a $30-60-90$ right triangle; since $BX=12$ then $BY=12\cos(60^{\circ})=6$ and the claim follows. I have not managed to finish yet, but this seemed important enough that I thought to post it while I work on the rest. $\endgroup$ Commented Dec 30, 2016 at 20:13

3 Answers 3

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A short solution from the construction suggested by @Fimpellizieri in the comments.

Let $X\in AP$ such that $\Delta CBX$ is isosceles, and $Y\in AB$ such that $XY\bot AB$. Then $\angle XBY=\angle CBA-\angle CBX=80^\circ-20^\circ=60^\circ$, and, hence, $|BY|=12\cdot\cos 60^\circ=6$, i.e. $Y=Q$.

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Now consider the quadrilateral $BQXP$. Since $\angle BQX=\angle BPX=90^\circ$, all four points lay on the same (circumscribed) circle. From the inscribed angles theorem it follows that $\angle BPQ=\angle BXQ$, but the latter is $90^\circ-60^\circ=30^\circ$. Finally, $\angle CPQ=90^\circ+30^\circ=120^\circ$.

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    $\begingroup$ Nice! I was having trouble coming up with a simple solution near the end. Yours is great. +1 $\endgroup$ Commented Dec 30, 2016 at 21:47
  • $\begingroup$ @Fimpellizieri It was only possible after your great construction, so it is more than on half your solution. $\endgroup$
    – A.Γ.
    Commented Dec 30, 2016 at 21:53
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Let us consider $\triangle APB$:

$$\angle ABP = 180° - (\angle APB + \angle PAB)$$ $$\angle ABP = 70°$$

Let us consider $\triangle BPQ:$

Let $\alpha=\angle BPQ$

$$ \dfrac{\sin{\alpha}}{BQ}= \dfrac{\sin{70°}}{QP}$$ $$ \sin{\alpha}= \dfrac{\sin{70°}}{QP}BQ$$ $$ \sin{\alpha}= 6\dfrac{\sin{70°}}{QP}$$

Now let us find $QP$

Let us consider $\triangle APQ:$

$$QP^2 = AP^2 + AQ^2 - 2\cdot AP\cdot AQ \cdot \cos(\angle PAQ)$$

Now let us find $AP \text{ and } AQ$

Let us consider $\triangle BPC:$

$$ \dfrac{\sin{\angle PBC}}{PC}= \dfrac{\sin{\angle PCB}}{BP}= \dfrac{\sin{\angle BPC}}{BC}$$ $$ BP = \dfrac{\sin{\angle PCB}}{\sin{\angle BPC}}\cdot BC$$ $$ BP = 11.82$$ $$ PC = \dfrac{\sin{\angle PBC}}{\sin{\angle PCB}}\cdot PB$$ $$ BP = 2.08$$

Let us consider $\triangle ABP\text{ to find } AP$

$$ \dfrac{\sin{\angle BAP}}{BP}= \dfrac{\sin{\angle ABP}}{AP}$$ $$ AP= \dfrac{\sin{\angle ABP}}{\sin{\angle BAP}}BP$$ $$ AP= 32.48$$

$P \in [AC] \Rightarrow AC = AP + PC \Leftrightarrow AC = 34.56$

Since $\triangle ABC \text{ is isosceles, then } AB = AC = 34.56$

$Q \in [AB] \Rightarrow AQ = AB - BQ \Leftrightarrow AQ = 28.56$

$$QP^2 = AP^2 + AQ^2 - 2 AP \cdot AQ \cdot \cos(\angle QAP)$$ $$QP = 11.28$$

Now let us consider $\triangle BPQ$

$$ \dfrac{\sin{\angle \alpha}}{BQ}= \dfrac{\sin{\angle QBP}}{QP}$$ $$ \sin{\alpha}= \dfrac{\sin{\angle QBP}}{QP}BQ$$ $$ \sin{\alpha}= \dfrac{1}{2} \Rightarrow \alpha = 30°$$

$$\angle CPQ = \angle CPB + \angle BPQ$$ $$\angle CPQ = 90° + 30°$$ $$\angle CPQ = 120°$$

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  • $\begingroup$ This solution is really huge -_- There should be some easier solution.. Btw. Nice (Y) (+1) $\endgroup$ Commented Dec 30, 2016 at 21:00
  • $\begingroup$ I already got solution from Fimpellizieri's comment :) $\endgroup$ Commented Dec 30, 2016 at 21:01
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Since $\angle BPC = 90^{\circ}$, it follows that $P$ lies on the circle whose diameter is $BC$. As such, let $D$ be the midpoint of $BC$, and let $E$ be a point inside $\Delta ABC$ such that $\Delta BDE$ is equilateral. Then $E$ also lies on the circle with diameter $BC$, so quadrilateral $BEPC$ is cyclic. Thus, $\angle BEP$ and $\angle PCB$ are supplementary, and since $\angle PCB = 80^{\circ}$, it follows that $\angle BEP = 100^{\circ}$. Now, $BC = 12$, so $BE = BD = \frac{1}{2}BC = 6$. Thus, $BE = 6 = QB$, so $\triangle QBE$ is isosceles. Since $$\angle QBE = \angle QBC - \angle EBC = 80^{\circ} - 60^{\circ} = 20^{\circ}$$ it follows that $\angle QEB = 80^{\circ}$. Thus, $\angle QEB$ and $\angle BEP$ are supplementary, so $E$ lies on line $QP$. As such, we have $$\angle CPQ = \angle CPE$$ and $\angle CPE = 120^{\circ}$, either by noting that it subtends an arc of $240^{\circ}$, or by noting that it is supplementary to angle $\angle EBC = 60^{\circ}$.

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