A Nice Geometry Problem | Math Olympiad | 2 Methods
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- Опубліковано 25 тра 2024
- A Nice Geometry Problem | Math Olympiad | 2 Methods
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At 13:58
Draw PQ perpendicular to CD at Q.
Since angle DPQ = 30 degrees, DQ/DP = 1/2
--> DQ = 2 --> BQ = 4+2 = 6
So Q is the mid-point of BC.
Draw BP,
Since PQ is perpendicular to BC and bisects BC, we will get
BP = CP, angle PBC = angle PCB = 30 degrees
--> angle BPC = 120 degrees.
So BP = CP = AP = 4*sqrt(3).
That means P is the center of circum-circle of ABC.
Angle BPC is 2*angle BAC.
But angle BPC = 120 degrees --> angle BAC = 60 degrees.
Angle ABC = 180 - angle BAC - angle ACB = 180 - 60 - 75 = 45 degrees.
Drop a perpendicular from A to BC and label the intersection as point M. ΔAMC is a 15°-75°-90° right triangle, for which long side AM is (2 + √3) times as long as short side CM. Let CM have length x, then AM has length x(2 + √3). Also, DM has length CD - CM = 8 - x. Note that ΔADM is a special 30°-60°-90° right triangle, so its long side AM is √3 times as long as short side DM, so x(2 + √3) = (√3)(8 - x). Expanding, 2x + x√3 = 8√3 - x√3. Then, 2x + 2x√3 = 8√3, 2x(1 + √3) = 8√3, x(1 + √3) = 4√3. Multiply both sides by (√3 - 1) and simplify to get x = 6 - 2√3. So MB = BD + DM = 4 + 8 - x = 12 - (6 - 2√3) = 6 + 2√3. AM = x(2 +√3) = (6 - 2√3)(2 + √3) = 12 - 4√3 +6√3 -2√3(√3) = 6 + 2√3. So, AM and MB have equal lengths, 6 + 2√3. Therefore, since
45 degrees
From the triangle on the right, create a 30-60-90 right triangle and a 45- 46- 90 right
triangle. Hence, from the 30-60-90 sides part of AD = 4
and from the 45-45-90 the remain length of AD = 4 sqrt 3
Hence, AD = 4 + 4sqrt 3 = 10.92820
To find the angles for triangle ABD, use 10.92820, 120 degrees and 4 and the
Law of Cosine. This yiedls a 45, 15, 120 triangle
Hence Answer = 45 degrees
I regret not commenting on this sooner but I think that both methods make way more sense than the comment section. The firs method just constructs two triangles a la the Law of Sines and is computed through trig identities as the ratios set equal. And the second method constructs right triangle in the first triangle contained and just uses basic cosine identity and then constructs a triangle that contrains the other triangle. I think that the second method is faster. I could use the first method as a way to practice remembering the right trig identities and force myself to be more careful.
Side length |AC| from sine law in triangle ACD
Side length |AB| from cosine law in triangle ABC
Angle measure ABC = theta from cosine law in triangle ABC
PD is 8*1/2=4. Drop a perpendicular from P to BC and call the intersection Q. Then DQ=1/2*PD=2; BQ=QC=6; BP=CP; angle PBQ=PCQ=30; triangle APC is isosc(45-90-45) so CP=AP=BP; angle BPD = 120-90=30; angle ABP= 0.5*30=15; angle ABD = angle ABP + angle PBD = 15 + 30 = 45. This involves virtually no calculation or trigonometry.
Even simpler method. Let the midpoint of BC be Q. Raise a perpendicular from Q and call the intersection with AD, P. BP=CP immediately follows.
DP=2*DQ=2*2=4=0.5*DC ; angle DPC= 90 angle DCP=30; angle PCA = 45; triangle APC is 45-90-45 isosc thus AP=CP=BP; P is the center of the circumscribing circle and the angle APC = 1/2 of anlge APC = .5*90=45
asnwer=45 isit
Yes.
AM/DM = tan60
AM/MC = tan75
DM/MC = tan60/tan75 and DM + MC = 8
2 unknowns 2 equation, which DM could be solved, and AM will be known,
Correction:DM/MC = tan75/tan60, not tan60/tan75. Your method is the method of choice in exam as DM can be found quickly using calculator and then tan(theta) = DM tan60/(4 + DM) = 1. I've checked the result. If you don't have a calculator, try tan(30+45)/tan60 to get DM = 4(sqrt3 + 2)/(sqrt3 + 1) using trigonometric identity and then get tan(theta) = 1.