Inter-related geometry problem
Sujeet Kumar
Q1.In a triangle ABC, point P is taken so that angle ABP=30 degree, angle PBC=70 degree and angle PCB=10 degree=angle PCA.Find angle BAP.
Q2.-In a triangle ABC, point P is taken so that angle ABP=70 degree, angle PBC=10 degree=angle PCB and angle PCA=30 degree.Find angle BAP.
In one of my previous mail I had mentioned that Dave Rusin had given a list of combination of angles
where point is taken inside the triangle and 6 formed angles are multiples of 10 degree.Combination
related to problem Q2 is [70,10,10,30,20,40] and this combination was absent in Dave Rusin's list.
Rusin's list contained 74 total combinations.There are total 43 interesting combinations which are non-trivial and non obvious.I have solved 42 of them, one case is still unsolved.We can form a problem if we give 4 angles out of 6.So with one combination we can form C(6,4)=15 problems.So we have total
of 43*15=645 , real bad problems.I think I am not wrong if I say that if we solve one of 15 then all 15
are solved.But if we try to find direct solution of all 15 then all have different type of solution.If some
person has got lot of free time then he can try to find direct solution of all 15.Even among 43 problems
we can find inter-relation with some simple construction as in above 2 problems.
I will write more in coming mails.
**No response is depressing.I hope I am doing something which has got some worth.**
Philippe 92
There is much much more to say about these "inter-dependance"
relations. I write today just about "directly equivallent" problems.
That is which result into exactly the same figure.
See attachement.
The figure shows two choices of 4 given angles in the 6.
The first is an ill-stated assignement :
Q. Let ABC a triangle and P inside with angle ABP = 70 deg,
PBC = PCB = 10 deg and PAC = 20 deg. Find angle BAP.
There are two non equivallent solutions, so that we can't decide
if the answer is 30 deg (point A) or 40 deg (point A') !
This was exactly the aim of my "A different geometry problem" :
To exhibit a specific choice of 4 angles among 6 with no unique
answer, so I had to append a specific additional condition.
To get the exact number of valid assignements (with unique solution)
for the same figure, we have to consider the way we may construct
the figure from the data.
Considering the four triangles, (ABC, APB, BPC, CPA), the solution
will be unique if two of them are fully defined, that is they each
contain two given or known angles.
This results into C(4,2) = 6 assignements with obvious construction
(not obvious solution, but just _constructing_ the figure)
[70,10,10,30,?,?]
[70,10,?,?,20,40]
[?,?,10,30,20,40]
[?,10,10,30,20,?]
[70,10,10,?,?,40]
[70,?,?,30,20,40]
But there is also the case when just one triangle is fully known,
and the two others are not, but with adjacent given angles.
For instance with given PBC, PCB, PAB and PAC, triangle PBC is fully
defined, but to construct A requires some effort.
Then A is constructible through the intersection of circles through
BP, substending known angle PAB, and through PC, substending known
angle PAC. As P is already common to these two circles, there is
only one solution for A (at most, with P inside ABC).
It is the second figure in the attachement.
This corresponds to valid assignements :
[?,10,10,?,20,40] (the one in the attachement)
[70,?,10,30,?,40]
[70,10,?,30,20,?]
Hence just 9 valid assignements instead of 15.
The six others are "ill-defined", as in the first example.
These result from just one known triangle (3 choices) and
"alternate" given angles in the two others. Hence 2*3 = 6 cases.
That's all for now, I won't discuss today of "indirect" equivallence
of two different figures being equivallent problems.
Merry Christmas.
Attachment available from http://mathforum.org/kb/servlet/JiveServlet/download/129-2223034-7342519-662920/notsame.gif
Sujeet Kumar
Here is solution of Q1.-
Construct an equilateral triangle MBC so that M is on the same side of BC as P.
Angle BAC=180-(30+70+10+10)=60 degree=angle BMC.
So, quadrilateral ABCM is cyclic.
Let BM and AC intersect at Q.
Angle PBQ=70-60=10 degree=angle QCP.
So, quadrilateral PBCQ is cyclic.
Angle PQB=angle PCB=10 degree.
Angle PQA=180-(angle BQC+angle PQB)=180-(100+10)=70 degree
Placement of point P in triangle ABQ corresponds to placement of P in triangle ABC of problem Q2.
Here is solution of problem Q2.-
Angle BAC=180-(70+10+10+30)=60 degree=angle BMC
So, quadrilateral ABCM is cyclic.
Angle MAC=angle MBC=60 degree
Angle AMB=angle ACB=40 degree
Angle BMP=30 degree
Let AC and BM intersect at T.
Angle TMP=30 degree=angle TCP
So, quadrilateral TMCP is cyclic.
Angle TPM=angle TCM=60-40=20 degree.
Now considering triangle AMP and placement of point T.
This configuration corresponds to one of our earlier
solved problems.
If Philippee is still reading it.In one of his mails
he had written that he will write more about indirect
equivalence of these problems.I am waiting for his words.