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Here, $\Delta ABC$ is an isosceles triangle, where, $AB = AC$. $P$ is such a point interior to triangle $\Delta ABC$ so that some angles are formed inside, as shown in the figure.

$\angle BAP = ?$

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    Which contest is this taken from? – 2017-01-16
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    It's from Bangladesh Mathematics Olympiad 2016 – 2017-01-16

3 Answers 3

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Construct the circle of center $A$ and radius $AB=AC$, and let $O$ be the point where line $CP$ intersects the circle again: we have $\angle BOC={1\over2}\angle BAC$ and $\angle OAB=2\angle BCO=60°$. It follows that $ABO$ is an equilateral triangle.

Construct now the circle of center $O$ and radius $OB=OA$ and observe that it passes through $P$ because $\angle BPA=150°$. We have then $\angle BAP={1\over2}\angle BOP={1\over4}\angle BAC$. From this equality it follows that $\angle BAP={1\over3}\angle PAC=13°$.

enter image description here

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    .....Excellent! – 2017-01-16
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    .....Very nice! – 2017-01-16
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    It's extra ordinary!!!! @Aretino, can you give me some tips, to be so skillful in brainstorming geometric constructions to solve, or if you could please suggest me some websites or books where I can find a lot of mathematical problems of this level (i.e. the one we are discussing here) And, Thank you so so much – 2017-01-16
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    @Aretino, I didn't understand the statement, " the circle of center OO and radius OB=OAOB=OA and observe that it passes through PP because ∠BPA=150°" – 2017-01-16
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    Any inscribed angle subtending larger arc $AB$ is half the corresponding central angle, that is half of 300°. Conversely, an angle of 150° subtending chord $AB$ (opposite to $O$) must be an inscribed angle, and its vertex lies on the circle. – 2017-01-16
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    As for the tips: learn to use GeoGebra and experiment with it. Drawing a diagram with GeoGebra, I realized that $\angle BAP$ is always one quarter of $\angle BAC$, no matter the size of $\angle BAC$. In addition, I used the properties of inscribed angles to construct point $P$, and that triggered some thoughts. – 2017-01-16
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    A relay of thoughts, that was brilliant technique – 2017-01-16
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An isosceles triangle has two equal sides as well as two equal angles. Thence triangles $\Delta ABP$ and $\Delta ACP$ have two sides in common. Applying the Sine Rule to both gives

$$2\sin(2t-51)\sin(141-t) = \sin(t)$$

where $t$ is $\gamma_1$ in John's diagram. This has solution $t = 34$, leading to $\angle BAP = 13^\circ$.

[ $2\sin(17^\circ)\cos(17^\circ) = \sin(34^\circ)$ ]

I suspect there's an easier way to get there, but can't see it. (Probably involving $\gamma_1 = 2\beta_2$)

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    The first equation is correct and the answer $13^\circ$ is correct also, but I am not following how you get from that equation to the conclusion that $t=34$. I do see how to get $t=34$ using the identity $2\sin(a)\sin(b)=\cos(a-b)-\cos(a+b)$, however. – 2017-01-16
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    Nevermind, I was misreading it. – 2017-01-16
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    I agree there must be an easier way to do this, perhaps using some arcane theorem from geometry. :) By the way, there is one other solution satisfying the three given angles where the solution is also $13^\circ$ but the triangle is not isosceles. – 2017-01-16
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    @David Hartley, how do we get $t = 13$, from the equation $2sin(2t-51)sin(141-t) = sin(t) ? $ – 2017-01-23
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    You don't :) 13 is the size of the angle BAP, as required. t is angle ACP. Since sin(141-t) = cos(51-t) we can use the identity 2sin(a)cos(b) = sin(a+b) + sin(a-b) to get sin(3t - 102) = 0 and so t = 34. – 2017-01-23
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We have the following six linear equations in six variables: $\alpha,\alpha_1, \beta,\beta_1\beta_2,\gamma_1$

  1. $\alpha_1+39+\beta_1+\beta_2+\gamma_1+30=180$
  2. $\alpha_1+\beta_2+150=180$
  3. $39+\beta+\gamma_1=180$
  4. $\alpha+\beta_1+30=180$
  5. $\alpha+\beta+150=360$
  6. $\beta_1+\beta_2=\gamma_1+30$

Giving the following system

\begin{align} \alpha && \alpha_1 && \beta && \beta_1 && \beta_2 && \gamma_1&&\\ 0 && 1 && 0 && 1 && 1 && 1 &&\quad 111\\ 0 && 1 && 0 && 0 && 1 && 0 && \quad30\\ 0 && 0 && 1 && 0 && 0 && 1 &&\quad 141\\ 1 && 0 && 0 && 1 && 0 && 0 &&\quad 150\\ 1 && 0 && 1 && 0 && 0 && 0 && \quad210\\ 0 && 0 && 0 && 1 && 1 && -1 &&\quad 30 \end{align}

Solving this by hand [ugh! wish I had a CAS] I get

  1. $\alpha=69+t$
  2. $\alpha_1=81-2t$
  3. $\beta=141-t$
  4. $\beta_1=81-t$
  5. $\beta_2=-51+2t$
  6. $\gamma_1=t$

In order for all six angle to be positive it must be the case that $\vert t-33\vert<7.5$.

triangle

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    I believe the third line in your matrix should have a $1$ in the $\gamma_1$ column. In addition, sum of lines $2$, $3$, and $4$ is the same as the sum of lines $1$ and $5$. – 2017-01-15
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    @AkivaWeinberger You are correct, I mis-typed the $\gamma_1$ value. And you are correct about the system being dependent. Thanks for pointing that out. I did not have energy left to decide. – 2017-01-15
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    It's just saying that the sum of the angles in the three subtriangles is the same as the sum of the angles in the big triangle plus the green stuff – 2017-01-15