# Math Olympic Question answer: Goal 2017 Junior Maths League training Team homework answer-1

Source: Internet
Author: User

Course Links: goal 2017 junior Math League training Team-1 (Zhao Yin)

1, right Angle $\triangle{abc}$, $AD$ is the hypotenuse on the high, $I _1, i_2$ is $\triangle{abd}, \triangle{acd}$ heart, verify: $B, C, I_2, i_1$ four points round.

Prove:

Consider proof $$\angle{bi_1i_2} + \angle{bci_2} = \angle{bi_1d} + \angle{di_1i_2} + \angle{bci_2} = 180^\circ.$$ easy to know $$\angle{BI_1D} = 180^\circ-{1\over2} (\angle{abd} + \angle{adb}) = 180^\circ-{1\over2} (180^\circ-\angle{bad}) =90^\circ + {1\over2} \angle{bad}.$$ on the other hand, $$\angle{bac} = \angle{i_1di_2} = 90^\circ,\ Ab:ac = i_1d:i_2d\ (\because\triangle{abd} \sim \tri ANGLE{CAD}).$$ so $$\triangle{abc} \sim \triangle{di_1i_2}\rightarrow \angle{abc} = \angle{di_1i_2},\ \angle{ACB} = \ angle{di_2i_1}.$$ this can be $$\angle{bi_1i_2} + \angle{bci_2} = \angle{bi_1d} + \angle{di_1i_2} + \angle{bci_2}$$ $$= 90^\circ + {1\over2}\angle{bad}+ \angle{abc} + {1\over2}\angle{acb}$$ $$= 90^\circ + {1\over2}\angle{bad}+ \angle{ABC} + {1\over2} \angle{bad}$$ $$= 90^\circ + \angle{bad}+ \angle{abc} = 180^\circ.$$

Q.E.D.

2, in the convex Pentagon $ABCDE$, if $\angle{abc} = \angle{ade}$, $\ANGLE{AEC} = \angle{adb}$, verify: $\angle{bac} = \angle{dae}$.

Prove:

Set $BD, the ce$ intersection is $F$. by $\ANGLE{AEC} = \angle{adb}$ $A, E, D, f$ Four points are all round.

Again by $\angle{abc} = \angle{ade} = \angle{afe}$ know $A, B, C, f$ Four points are all round.

This can be $\angle{bac} = \ANGLE{BFC} = \angle{dfe} = \angle{dae}$.

Q.E.D.

3, in the circle of the string $AB, ac$, $\angle{bac}$ of the split line round at the point $D$. Over $D$ for $DE \perp ab$ in $E$, verify: $AE = \displaystyle{1\over 2} (AB + AC)$.

Prove:

Consider proving $2ae = AB + ac$.

By the nature of the angular line and the $CD = bd$, constructs the congruent triangle, makes $DF \perp ac$ (or its extension cord) in $F$.

Easy to $\triangle{dfc}\cong\triangle{deb}$ and $\triangle{dfa}\cong\triangle{dea}$, so $2ae = AE + AF = AB + ac$.

It should be noted that the case should be explained when the $B, e$ coincident. Ishi at this time $AD$ for the diameter of the circle, the conclusion is still established.

Q.E.D.

4, in the Triangle $ABC$, $\ANGLE{ACB} = 45^\circ$, $D$ for $AC$ on a point and $\angle{adb} = 60^\circ$, $AB$ cut $\triangle{bcd}$ circumscribed circle in $B$, verify: $AD:D C = 2:1$.

Prove:

Because of the $AD, the dc$ in the same line, considering the parallel cut theorem and the parallel relationship to solve.

Connect $OB$ to create a right angle, and then by $\ANGLE{ACB} = 45^\circ$ Consider recreating a $90^\circ$: $$\angle{bod} = 2\ANGLE{BCD} = 90^\circ.$$ thus found $AB \parallel Od\rightarrow AD:DC = be:ec$. Next you can consider proving $BE: EC = 2:1$.

Calculated by $\triangle{abc}\sim\triangle{adb}$ and simple angle: $$\angle{abc} = \angle{adb} = 60^\circ\rightarrow \angle{OBC} = 30^\ Circ\rightarrow be = 2oe,$$ $$\angle{dbc = 15^\circ}\rightarrow \angle{doc} = 30^\circ\rightarrow EC = OE.$$ $BE = 2C E\rightarrow AD = 2dc$.

Q.E.D.

5, set $ABCD$ is rounded inside the quadrilateral, diagonal $AC$ with $BD$ in $X$, from $X$ for $AB, BC, CD, da$ perpendicular, perpendicular respectively for $A ^\prime, B^\prime, C^\prime, d^\prime$, Verification: $A ^\prime b^\prime + c^\prime d^\prime = a^\prime d^\prime + b^\prime c^\prime$.

Prove:

A known vertical relationship is available $A ^\prime, B, B^\prime, x$ four points, with a diameter of $BX$. Then by the sine theorem there is $${bx\over\sin90^\circ} = {a^\prime b^\prime \over \sin\angle{a^\prime bb^\prime}}\rightarrow {A^\prime B^\ Prime\over BX} = \sin\angle{a^\prime bb^\prime}.$$ on the other hand, in $ABCD$ circumscribed circle also has $${ac\over d} = \sin\angle{a^\prime bb^\prime},  d of which is ABCD  circumscribed circle diameter. The  A ^\prime b^\prime = {Bx\cdot Ac\over d}.$$ similarly can be C ^\prime d^\prime = {Dx\cdot Ac\over d}.$$therefore,  A ^\prime B^\prim E + c^\prime d^\prime = bd\cdot{AC \over d}.$$ similarly available B ^\prime c^\prime + a^\prime d^\prime = ac\cdot{BD \over d}.$$ie  A^\prime B^\prime + c^\prime d^\prime = a^\prime d^\prime + b^\prime c^\prime. Q.E.D. 6, quadrilateral ABCD  within the circle, the center of the other circle O  on the side AB , and tangent to the remaining three sides, verify: AD + BC = ab. Prove: Considered by the proof conclusion to be AD, the bc translates to AB , and the BE is intercepted on AB  = bc, which proves that the target is AE = ad. by A, B, C, d four points round and tangent properties of$$\angle{bec} = \angle{bce} = {1\over2} (180^\circ-\angle{ebc}) = {1\OVER2}\ANGLE{ADC} = \angl e{odc},$$therefore O, E, C, d Four points are all round. This can be$$\angle{aed} = \angle{ocd} = {1\over 2}\angle{bcd} = {1\over2} (180^\circ-\angle{a}), $$is proof ae = ad. Q.E.D. 7, the two circles with each other in the D , the line cut a round in A , the other round in the B, C  Two, verify: A  to a straight line BD, cd distance equal. Prove: The distance between the points and the two lines (line segments) is equal, which can be linked to the nature of the angular divide line. Extending the CD  round in E , consider proving AD  is the \angle{bde} of the Horn of the split line. Over point D  make two rounds of Gongsche AC  F ,$$\angle{adb} = \ANGLE{ADF} + \angle{bdf} = \angle{daf} +\ANGLE{DCF} = \angle{ade}. that proves $Ad$ is the $\angle{bde}$ of the horn.

Q.E.D.

Scan attention to "Austrian mathematics Garden" public number (Id:aoshu_xueyuan)

Math Olympic Question answer: Goal 2017 Junior Maths League training Team homework answer-1

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