Find The Angle
In △ A B C , ∠ B A C = 5 0 ∘ . A circle Γ is tangent to side A B and A C at points P , Q repectively. Given that P Q ∥ B C , find ∠ A C B in degrees.
Details and assumptions
- This problem is inspired by an Iran TST geometry problem.
The answer is 65.
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2 solutions
I solved it same way by realizing this is an isosceles triangle but didn't do all those steps.
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I was just listing them all out to show my thought process. I too solved it as soon as I looked at it. Did you go to State MATHCOUNTS? I know that a lot of people at your school did, like @William Cui .
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Hmm I don't know anyone at Takoma named Mardokay Mosazghi, but apparently he/she lives in Rockville, which is pretty close to where I live.
By the way, how did you know I went to State MathCounts? :P
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@William Cui – Because you're going to Nationals for Maryland. I'm kind of like one of the ESPN announcers who knows the game in and out, and the players, and stuff like that, but can't actually play. I didn't do too well at States (I'm in VA which is really tough), but I'm a professional "analyzer" of competitions. Let's just say... I've done my homework. :D Yeah dude, what did you get at States?
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@Finn Hulse – Oops I was sick at States so I did subpar (I got a 40 lol).
Yeah VA is pretty good this year (AK > me at JHMT lol, Franklyn is good at hugging me but better at Countdown - he beat me at Carderock CD finals, and Will Sun is pro at cubing and math lol. Joshua is probably also OP but idk many people from Rocky Run.)
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@William Cui – Oh yeah. They're all super good though. :D
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@Finn Hulse – Yeah.
I know most of the VA team pretty well, considering that I don't see them on a daily basis (at least not in person).
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@William Cui – Wait... describe how you know them... :O
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@Finn Hulse – Well I went to a bunch of competitions w/ them so I talk to them and also on AoPS.
@William Cui – Yeah you don't know me, I wasn't in the magnet and I aslo didnot participate in math counts but apparently I know you i attend MBHS I am also inn the math team but I did not participate in math clubs or anything last year.. @William Cui and @Finn Hulse
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@Mardokay Mosazghi – How did you not make the magnet program?
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@Finn Hulse – I did-not make it because the application was turned in in 5th grade and I came to US in 7th grade also my English is so bad.
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@Mardokay Mosazghi – Dude that sucks. What country are you from? I have a similar problem where I'm a Level 5 Algebra person on Brilliant and I'm in Algebra 1 at my school which sucks. :O
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@Finn Hulse – Eritrea, probably i am the only one here who uses brilliant and let delete those replys I think Sreejato Bhattacharya is going to be mad.
Another way to prove △ A B C is isoceles:
Since ∠ Q P B = 1 8 0 ∘ − ∠ Q P A = 1 8 0 ∘ − ∠ A Q P = 1 8 0 ∘ − ∠ Q C B , trapezoid Q P B C is cyclic. But every cyclic trapezoid must be isoceles, so we must have B P = Q C , and hence A B = A P + P B = A Q + Q C = A C .
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Yes, but you end up proving just one of the laws of circles.
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Nope, that's what you did. ;) It is well known that any cyclic trapezoid must be isoceles, and that's what you proved. Whatever, nice solution!
By tangent property
AP = AQ
BAC = PAQ = 50
so AQP = QPA = 65
PQC = BPQ = 115
as quadri PQCB a trapezium
PQC + QCB = 180
QCB = ACB = 65
can you send the draw of problem
Fantastic problem @Sreejato Bhattacharya ! Finally one I can even understand! Okay, here's how I did it. First realizing that by PP, triangle Q A P must be isosceles given that A Q and A P are truly tangent to the circle. By this same logic, triangle A B C also is isosceles. Because they share a common angle, an the bases across from that angle are parallel, it is apparent that A Q P ∼ A B C . Now, given what we've learned above, angles A Q P and A P Q must both be the same. Thus, they are both 2 1 8 0 − 5 0 = 6 5 . Thus, by the similarity proved before, angle A C B must also be 6 5 .