10
$\begingroup$

$\triangle ABC$ 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

$\endgroup$
12
  • $\begingroup$ What is the way you need to solve this? $\endgroup$ Dec 30 '16 at 17:52
  • $\begingroup$ @Fimpellizieri Just without using Calculator. $\endgroup$ Dec 30 '16 at 17:52
  • $\begingroup$ the answer is $$120^{\circ}$$ $\endgroup$ Dec 30 '16 at 18:15
  • $\begingroup$ I found that same answer using laws of cosines but this is terribly long. $\endgroup$ Dec 30 '16 at 19:16
  • 3
    $\begingroup$ 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. $\endgroup$ Dec 30 '16 at 20:13
8
$\begingroup$

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. $Y=Q$.

enter image description here

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, $\angle CPQ=90^\circ+30^\circ=120^\circ$.

$\endgroup$
2
  • 1
    $\begingroup$ Nice! I was having trouble coming up with a simple solution near the end. Yours is great. +1 $\endgroup$ Dec 30 '16 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.
    Dec 30 '16 at 21:53
2
$\begingroup$

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 $\triangle BPQ$

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

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

$\endgroup$
2
  • $\begingroup$ This solution is really huge -_- There should be some easier solution.. Btw. Nice (Y) (+1) $\endgroup$ Dec 30 '16 at 21:00
  • $\begingroup$ I already got solution from Fimpellizieri's comment :) $\endgroup$ Dec 30 '16 at 21:01
1
$\begingroup$

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 $\angle EBC = 60^{\circ}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.