# Geometry Exam Paper

Junior Exam J2

Each question is worth 7 marks.

Time: 4 hours

No books, notes or calculators allowed.

Q1

Let $AB$ be the diameter of circle $\Gamma$. Let $C$ be a point on line $AB$ outside $\Gamma$. A tangent from $C$ touches $\Gamma$ at point $N$. The bisector of $\angle ACN$ intersects segments $AN$ and $BN$ at points $P$ and $Q$, respectively.

Prove that $PN$ = $QN$.

Q2

Let $ABCD$ be a trapezium whose parallel sides are $BC$ and $AD$. Let $O$ be the intersection of the trapezium's diagonals $AC$ and $BD$. Suppose further that $CD = AO$, $BC = DO$ and that $CA$ is the bisector of $\angle BCD$.

What is the value of $\angle ABC$?

Q3

Point $P$ is situated on the hypotenuse $AB$ of right-angled triangle $ABC$, and satisfies

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

Calculate $BC : AC : AB$.

Q4

Points $D$ and $E$ lie on sides $AC$ and $BC$, respectively, of triangle $ABC$. It is known that $\angle BDE = 90^{\circ}$ and $AD = AB = BE$.

Prove that $AB + AC = 2BC$.

Q5

Let $O$ be the circumcentre of acute triangle $ABC$. A circle passing through points $B$, $O$ and $C$ intersects line $AB$ for a second time at point $D$ and intersects line $AC$ for a second time at point $E$.

Prove that lines $AO$ and $DE$ are perpendicular.

Note by Sharky Kesa
4 years, 9 months ago

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Q5 (Not completed, need a sleep)

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

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

$B\hat{D}O = B\hat{C}O = B\hat{E}O = O\hat{D}C$ and

$C\hat{E}O = C\hat{B}O = B\hat{C}O = B\hat{E}O$. (extra: everything above is equal!)

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

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

- 4 years, 9 months ago

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

Q5: Let the perpendicular bisector of $AC$ intersect $AB$ at $D'$. We have $\measuredangle BD'C=\measuredangle AD'C=2\measuredangle BAC=\measuredangle BOC$ meaning $BCOD'$ cyclic so $D'=D$. Similarly $E'=E$. Now in $\triangle ADE$ we have $O$ orthocenter, the result thus follows. (Angles are directed mod $\pi$)

- 4 years, 9 months ago

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

- 4 years, 2 months ago

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

- 3 years, 9 months ago

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.

- 3 years, 9 months ago