Geometry Exam Paper

Junior Exam J2

Each question is worth 7 marks.

Time: 4 hours

No books, notes or calculators allowed.

Note: You must prove your answer.

Q1

Let ABAB be the diameter of circle Γ\Gamma. Let CC be a point on line ABAB outside Γ\Gamma. A tangent from CC touches Γ\Gamma at point NN. The bisector of ACN\angle ACN intersects segments ANAN and BNBN at points PP and QQ, respectively.

Prove that PNPN = QNQN.

Q2

Let ABCDABCD be a trapezium whose parallel sides are BCBC and ADAD. Let OO be the intersection of the trapezium's diagonals ACAC and BDBD. Suppose further that CD=AOCD = AO, BC=DOBC = DO and that CACA is the bisector of BCD\angle BCD.

What is the value of ABC\angle ABC?

Q3

Point PP is situated on the hypotenuse ABAB of right-angled triangle ABCABC, and satisfies

PB:PC:PA=1:2:3PB : PC : PA = 1 : 2 : 3

Calculate BC:AC:ABBC : AC : AB.

Q4

Points DD and EE lie on sides ACAC and BCBC, respectively, of triangle ABCABC. It is known that BDE=90\angle BDE = 90^{\circ} and AD=AB=BEAD = AB = BE.

Prove that AB+AC=2BCAB + AC = 2BC.

Q5

Let OO be the circumcentre of acute triangle ABCABC. A circle passing through points BB, OO and CC intersects line ABAB for a second time at point DD and intersects line ACAC for a second time at point EE.

Prove that lines AOAO and DEDE are perpendicular.

Note by Sharky Kesa
4 years, 9 months ago

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Q5 (Not completed, need a sleep)

Construct DO,EO\overline{DO},\overline{EO} intersect AC,AB\overline{AC},\overline{AB} at point G,FG,F respectively. (sorry for someone with OCD. (not in the problem!))

We know that BDCO,CEBO\square BDCO, \square CEBO are concyclic, we get

BD^O=BC^O=BE^O=OD^CB\hat{D}O = B\hat{C}O = B\hat{E}O = O\hat{D}C and

CE^O=CB^O=BC^O=BE^OC\hat{E}O = C\hat{B}O = B\hat{C}O = B\hat{E}O. (extra: everything above is equal!)

** Need to prove OFAB\overline{OF} \perp \overline{AB} and OGAC\overline{OG} \perp \overline{AC}.

Consider AEF\triangle AEF; by definition of orthocenter, we get OAEF\overline{OA} \perp \overline{EF}.

Samuraiwarm Tsunayoshi - 4 years, 9 months ago

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Q1: By Alternate Segment theorem πANC=NBA=πCBN\pi-\angle ANC=\angle NBA=\pi-\angle CBN so we deduce CBQPNC\triangle CBQ\sim\triangle PNC. Hence CPN=BQC=NQP\angle CPN=\angle BQC=\angle NQP implying PN=QNPN=QN.

Q5: Let the perpendicular bisector of ACAC intersect ABAB at DD'. We have BDC=ADC=2BAC=BOC\measuredangle BD'C=\measuredangle AD'C=2\measuredangle BAC=\measuredangle BOC meaning BCODBCOD' cyclic so D=DD'=D. Similarly E=EE'=E. Now in ADE\triangle ADE we have OO orthocenter, the result thus follows. (Angles are directed mod π\pi)

Jubayer Nirjhor - 4 years, 9 months ago

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a circle has two types of tangents.which one is the largest?

diali kundu - 4 years, 2 months ago

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Finally cracked question 4. My solution is as follows : What we have to prove is equivalent to proving that DC=2EC. Thus we proceed as follows. Drop a perpendicular from A to BD and let it meet BD at G and BC at F. Notice that ABD is an isosceles triangle and hence BG=GD. Observe that GF is parallel to DE. Hence BF=FE. Now we have triangle CED similar to triangle CFA. Thus, CD/DA=CE/EF. Use the fact that AD=2EF. Hence obtain that DC=2.EC Hence proved

Shrihari B - 3 years, 9 months ago

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I am also done with question 3. Denote by O the midpoint of AB. Notice that OP=OR=R/2 , PC=R and hence apply Appolonius Theorem on triangle OBC with the median as PC. We get a(i.e. BC)=R(1.5)^0.5 We also know that c=2R. Hence by Pythagoras theorem we can find b and hence we can get the ratio of the sides.

Shrihari B - 3 years, 9 months ago

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