Triangle PQR is entirely within equilateral triangle ABC and is positioned such that angle ABP = angle = 15°, angle CBR = angle BCQ = 20° and angle CAR = angle ACP = 25°
What are the angles of triangle PQR?
(Sorry, you have to draw your own picture as part of the solution.)
Dave:
Did you really mean to pose the problem as posted or as :
> Triangle PQR is entirely within equilateral triangle ABC and is positioned such that angle ABP = angle BAQ = 15°, angle CBR = angle BCQ = 20° and angle CAR = angle ACP = 25°
Yep.
> Yep.
Obviously (I think), the most subtle part of that problem is why the directions of the sides of the triangles should differ by such regular angles. Nice problem.
Kent has drawn it for you, now all you have to do is solve it.
Here's one method of solution:
Let A-B = B-C = C-A = 1
Solve angles A-R-B, C-P-B, A-Q-C
Use law of sines to solve C-P, A-Q and A-R
Let P2-P, Q2-Q, R2-R be perpendicuar to A-C
Solve A-P2, A-Q2, A-R2, P2-P, Q2-Q, R2-R
Compute angles of P-R, R-Q, Q-P relative to A-C
Solve interior angles of P-R-Q
Your penultimate line should be "Compute sides of..."
> Your penultimate line should be "Compute sides of..."
Except it shouldn't be necessary to actually compute the side lengths of the inner triangle to work the problem, right?
A Hint: the angles are 75,45 and 60 degrees exactly. Solved 2 different ways as a check.
JOHN NOLTON
Tombstone, AZ
Brute Force Solution - step by step
For visual convenience, refer diagrams by Kent McMillan
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[pre]1. for computational convenience, let sides of equilateral triangle ABC = 1,
ie (AB) = 1, (BC) = 1 and (AC) = 1
2. form triangles
a. AQC
b. ARB
c. CPB
3. determine all angles in triangles AQC, ARB and CPB
example for triangle AQC:
(QAR) = 60 - [(QAB) + (RAC)]
(QAC) = (QAR) + (RAC)
(PCQ) = 60 - [(ACP) + (QCB)]
(ACQ) = (PCQ) + (ACP)
(AQC) = 180 - [(QAC) + (ACQ)]
determine angles for triangles ARB and CPB similarly
4. using sine law
a. determine (AQ)
sin(AQC)/1 = sin(ACQ)/(AQ)
(AQ) = sin(ACQ)/sin(AQC)
b. determine (AR)
sin(ARB)/1 = sin(RBA)/(AR)
(AR) = sin(RBA)/sin(ARB)
a. determine (PB)
sin(CPB)/1 = sin(PCB)/(PB)
(PB) = sin(PCB)/sin(CPB)
5. determine point position coordinates relative to A
a. Qx = (AQ)cos(QAB) .. Qy = (AQ)sin(QAB)
b. Rx = (AR)cos(RAB) .. Ry = (AR)sin(RAB)
c. Px = 1 - (PB)cos(ABP) .. Py = (PB)sin(ABP)
6. using Pythagorean Theorem, determine the triangle sides
a. (PQ) = sqrt[(Qx - Px)^2 + (Qy - Py)^2]
b. (QR) = sqrt[(Qx - Rx)^2 + (Qy - Ry)^2]
c. (PR) = sqrt[(Rx - Px)^2 + (Ry - Py)^2]
7. using cosine law, determine two angles
a. (RPQ) = acos[((QR)^2 - (PR)^2 - (PQ)^2)/(-2(PR)(PQ)]
a. (PQR) = acos[((PR)^2 - (QR)^2 - (PQ)^2)/(-2(QR)(PQ)]
8. determine angle the remaining angle
(PRQ) = 180 - (RPQ) - (PQR)[/pre]
Note: Hopefully, there are no mistakes in the typing. It takes more time to type the solution than it does to complete it.
Also, I think there may be a more elegant solution using circles, inscribed and subtended angles, or tangents and secants, etc. It's Christmas and there's other matters to attend. Maybe later.
And the answer is...(well, one anyway)
