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 be the diameter of circle . Let be a point on line outside . A tangent from touches at point . The bisector of intersects segments and at points and , respectively.
Prove that = .
Q2
Let be a trapezium whose parallel sides are and . Let be the intersection of the trapezium's diagonals and . Suppose further that , and that is the bisector of .
What is the value of ?
Q3
Point is situated on the hypotenuse of right-angled triangle , and satisfies
Calculate .
Q4
Points and lie on sides and , respectively, of triangle . It is known that and .
Prove that .
Q5
Let be the circumcentre of acute triangle . A circle passing through points , and intersects line for a second time at point and intersects line for a second time at point .
Prove that lines and are perpendicular.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
Q5 (Not completed, need a sleep)
Construct DO,EO intersect AC,AB at point G,F respectively. (sorry for someone with OCD. (not in the problem!))
We know that □BDCO,□CEBO are concyclic, we get
BD^O=BC^O=BE^O=OD^C and
CE^O=CB^O=BC^O=BE^O. (extra: everything above is equal!)
** Need to prove OF⊥AB and OG⊥AC.
Consider △AEF; by definition of orthocenter, we get OA⊥EF.
Q1: By Alternate Segment theorem π−∠ANC=∠NBA=π−∠CBN so we deduce △CBQ∼△PNC. Hence ∠CPN=∠BQC=∠NQP implying PN=QN.
Q5: Let the perpendicular bisector of AC intersect AB at D′. We have ∡BD′C=∡AD′C=2∡BAC=∡BOC meaning BCOD′ cyclic so D′=D. Similarly E′=E. Now in △ADE we have O orthocenter, the result thus follows. (Angles are directed mod π)
a circle has two types of tangents.which one is the largest?
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
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.