An algorithm for composition of inverse problems with exclusive disjunction as a logical structure in the conclusion
aa r X i v : . [ m a t h . HO ] N ov AN ALGORITHM FOR COMPOSITION OF INVERSE PROBLEMSWITH EXCLUSIVE DISJUNCTION AS A LOGICAL STRUCTURE INTHE CONCLUSION
JULIA NINOVA AND VESSELKA MIHOVA
Abstract.
By suitable examples we illustrate an algorithm for composition of inverseproblems. introduction Let t, p , p , r be given statements. We deal with a generalization of the formal logicalrule [3] ( p → r ) ∧ ( p → r ) ⇔ p ∨ p → r. In [1] and [2] we prove, clarify and use the equivalence( ∗ ) ( t ∧ p → r ) ∧ ( t ∧ p → r ) ⇔ t ∧ ( p ∨ p ) → r. It gives an algorithm for composition of inverse problems with a given logical structurethat is based on the steps below.- Formulating and proving generating problems with logical structures of the state-ments as those at the left hand side of (*).- Formulating a problem with a logical structure of the statement t ∧ ( p ∨ p ) → r. - Formulating an inverse problem with a logical structure t ∧ r → p ∨ p .A technology for composition of equivalent and inverse problems is appropriate for trainingof mathematics students and teachers.In what follows we illustrate the above algorithm by two groups of problems.2. Group of problems I
The logical statements used for the formulation of the problems in this group are t : { In the plane are given a △ ABC , a straight line g ∋ C and an arbitrary point M ∈ g, M = C. } p : { The straight line g is a median in △ ABC. } p : { g k AB } r : { S △ AMC = S △ BMC } Problem 2.1.
Let in a △ ABC the straight line g passes through C and is a median. Provethat for any point M on g ( M = C ) the triangles △ AM C and △ BM C are equal in area.
Mathematics Subject Classification.
Primary 51F20, Secondary 51M15.
Key words and phrases.
Inverse problems, logical structure, exclusive disjunction.
JULIA NINOVA AND VESSELKA MIHOVA
This problem has a logical structure t ∧ p → r . Proof.
Since the straight line g passes through C and is a median in a △ ABC , then AB ∩ g = ∅ . Let AB ∩ g = D (fig. 1). Hence AD = BD .Let h and h be the altitudes in △ BDC and △ ADC through the vertices B and A respectively. Then S △ AMC S △ BMC = CM.h CM.h = h h = M D.h M D.h = S △ AMD S △ BMD = 1 . (cid:3) Problem 2.2.
Let in △ ABC the straight line g passes through C and is parallel to thestraight line AB . Prove that for any point M on g ( M = C ) the triangles △ AM C and △ BM C are equal in area.
This problem has a logical structure t ∧ p → r .The proof follows immediately from Figure 2.According to (*) and the generating problems t ∧ ( p ∨ p ) → r. Since the statements p and p are mutually exclusive, then p ∨ p ⇔ p ⊻ p . Under thiscondition problems with structures t ∧ ( p ∨ p ) → r and t ∧ ( p ⊻ p ) → r are equivalent.We formulate and solve the following inverse problem with a logical structure t ∧ r → p ⊻ p . Problem 2.3.
Let the point M in the plane of △ ABC is such that △ AM C and △ BM C exist and are equal in area. Prove that the straight line CM either cuts the side AB of △ ABC at its middle point or is parallel to AB . LGORITHM FOR COMPOSITION OF INVERSE PROBLEMS WITH EXCLUSIVE DISJUNCTION 3
Proof.
The triangles △ AM C and △ BM C have a common side CM and are equal inarea. Hence the altitudes to their common side are equal, i. e. the vertexes A and B areequidistant from CM .There are two possibilities for the location of the points A and B with respect to thestraight line CM .- If A and B are situated at opposite sides of CM (fig. 1), then the point D = AB ∩ CM is the middle point of the side AB ( △ AA D ∼ = △ BB D ) and the straight line CM is the median of △ ABC through C .- If A and B are situated at one and the same side of CM (fig. 2), then they lie on astraight line parallel to CM , i. e. CM k AB . (cid:3) Group of problems II
The logical statements used for the formulation of the problems in this group are t : { Let
ABCD be a quadrilateral with AC ∩ BD = O. } p : { AB k CD } p : (cid:26) ABCD = 1 (cid:27) r : (cid:26) AOOC = BOOD = λ (cid:27) Problem 3.1.
Let
ABCD be a trapezium with AC ∩ BD = O. Prove that
AOOC = BOOD = λ, λ = 1 . This problem has a logical structure t ∧ ( p ∧ p ) → r . Proof.
Since △ AOB ∼ △
COD (fig. 3) and AB = CD , then AOOC = BOOD = ABCD = λ, λ = 1 . (cid:3) Problem 3.2.
Let
ABCD be a parallelogram with AC ∩ BD = O. Prove that
AOOC = BOOD = λ, λ = 1 . JULIA NINOVA AND VESSELKA MIHOVA
This problem has a logical structure t ∧ ( p ∧ p ) → r . Proof.
Since AB = CD , the proportion AOOC = BOOD = ABCD = λ, λ = 1follows from the congruence △ AOB ∼ = △ COD (fig. 4). (cid:3)
According to (*) and the generating problems t ∧ (( p ∧ p ) ∨ ( p ∧ p )) → r. Since the notion trapezium is not generic with respect to the notion parallelogram, thenproblems with logical structures t ∧ (( p ∧ p ) ∨ ( p ∧ p )) → r and t ∧ (( p ∧ p ) ⊻ ( p ∧ p )) → r are equivalent.We formulate and solve the following inverse problem with a logical structure t ∧ r → ( p ∧ p ) ⊻ ( p ∧ p ) . Problem 3.3.
Let
ABCD be a quadrilateral with AC ∩ BD = O. Prove that if
AOOC = BOOD = λ, then the quadrilateral is either a trapezium or a parallelogram.Proof. From the given proportionality and the equality ∠ AOB = ∠ COD we concludethat △ AOB ∼ △
COD . Hence
ABCD = λ , ∠ OAB = ∠ OCD and ∠ OBA = ∠ ODC ,which implies the straight lines AB and CD are parallel.There are two possibilities for the ratio coefficient λ : either λ = 1 or λ = 1.- If λ = 1 the quadrilateral ABCD is a trapezium.- If λ = 1 the quadrilateral ABCD is a parallelogram. (cid:3)
LGORITHM FOR COMPOSITION OF INVERSE PROBLEMS WITH EXCLUSIVE DISJUNCTION 5
References [1] J. Ninova, V. Mihova (2014),
Composition of inverse problems with a given logical structure , to appearin Annuaire Univ. Sofia, Fac. Math. Inf.[2] V. Mihova, J. Ninova (2014),
On a generalization of criteria A and D for congruence of triangles , toappear in Annuaire Univ. Sofia, Fac. Math. Inf.[3] A. A. Stoljar (1971),
Logicheskoe vvedenie v matematiki , Vishejshaja shkola, Minsk., Vishejshaja shkola, Minsk.