Math is all about ideas. Solving a problem in an Olympiad involves two steps: solving the problem (or progress) and writing it down.
It is therefore important to be able to express our ideas well. Additionally, one should be able to convey their ideas in a way that is easily understood by others. You need to keep in mind that the people who read your solutions or proofs are readers and not mind readers. And sometimes, they can not guess what you are thinking. So try to convey your thoughts through your answer sheet as clearly as you can. If you gained any non-trivial information during the process of solving a problem, write it in the answer sheet if you feel it helps prove the statement.
The way you write your solutions and the number of details you skip will depend on the level of the exam you are taking. I would, for example, write some minute details in RMO that I would skip at the INMO level. Also, this post is for students who are completely new to olympiads and are trying to write olympiads of RMO/INMO level or basic proof-writing contests.
Another side note: Writing solutions for math camps application, a math course or research is very different from writing solutions in contests like TSTs or greater levels. Proof in a contest is just convincing the grader that you understand everything and the solution is complete.
In an application for math camp, research, etc., you are judged for your approach and thinking process. In a math camp, for instance, it is very advisable to write down your ideas as well as your solutions. We tend to use out-of-the-box ideas and think about the problem for days; we write down every single step we take and sometimes we use different materials, like colour pens, grids, and codes! (or at least I do) In another blog post, we will talk about math camps! Please stay tuned!
Another side note ( I promise this is the last): It is very critical to know what logically constitutes the proof. Moreover, even though this post might focus a lot on offering nice, beautiful solutions, you should remember that writing mathematically and logically correct solutions is much more important. Additionally, it is likely that if you are losing marks for a solution, it is because you made a mathematical error or you fake-solved it, rather than because of your ineffective writing. A write up which is readable, mathematically correct, and has no major details left out deserves to get full marks. (Thanks to Pranjal for helping!)
As stated above, a solution simply needs to be readable and neat in a contest. Below are some tips you could use in order to make your solution neater, organized, and loved by the reader. In general, it is also a wise practice to use all the time. Please do not feel overwhelmed by this post or scared by solution writing.
Contents:
- Solve the problem / get all possible ideas and progress (ft Atul)
- Plan how to write a solution/ progress
- Alignment and spacing
- Using asterisk or roman numerals
- Case works
- Handwriting
- Geometry diagrams
- How much detail to include
- General Tips ( ft Aatman Supkar)
Tip 0: Solve the problem / get all possible ideas and progress( ft Atul)
This point is so important we'll emphasise it at the beginning - SOLVING PROBLEMS IS MORE IMPORTANT THAN WRITING SOLUTIONS WELL.
Writing up solutions properly is a secondary skill. It only helps in making your solution easier to convey to someone reading it (or in a contest, the grader) and as long as it's legible and mathematically correct, you "should not" lose any marks on it. There are people with not great handwriting or writeups who have performed very well in even international contests. You should therefore focus on solving the problems in a contest instead of spending time trying to perfect your write-up. Although you should give yourself sufficient time, don't overdo it. We will have a post in the future here on "How to approach a problem" for that.
As Atul says, "It's more important to be able to solve problems than being able to write perfectly. A poorly written writeup may cost you maybe a point or two, but if you don't solve it, you'll get 0."
Tip 1: Plan how to write a solution/ progress
It's nice to have a plan before you start writing the solution. It's said that the type of solution you write reflects your thought process. A vague, unorganized solution reflects a vague thought process. After solving a problem, you can simply draw a tiny box and sketch an extra tiny short outline of your solution. Include the terms, points, intersections, processes, etc you’ll need to define, and the order in which you will write up the most crucial parts of your solution. (Note: you can draft the plan inside your brain too if you feel making a tiny box is too much work). Anyways, either of these processes has two benefits:
- It ensures that we don’t skip anything and that we put our steps/thoughts/progress in an order that’s easy to follow for the reader.
- It also helps us verify our solution (for example checking if it is a fake solution)
Now, planning before on how to write is really helpful. Since $$ \text{better planning} \implies \text{ fewer scribbles and neat work } \implies \text{ graders happy }$$ $$ \implies \text{ you get deserving marks } \implies \text { you happy } $$
Try to write the solution in such a way that the grader should get to know what you are thinking and why you are thinking so. Hence, it's a good practice to organize our solutions using claims and lemmas.
Benefits of organizing your solution using claims and lemmas:
- It organizes the solution so nicely, helps you get a clear idea, and creates a "step by step" solution
- Moreover, if we had to refer to a claim, we can simply number the claims and then refer ( it's not at all messy)
- It helps the reader understand the structure you are approaching
- Very helpful for getting partials
- A quick skim through the proof gives the reader an idea of whether the proof is right or wrong ( helping them save a lot of time)
For example, we present two solutions for RMM 2019 P1 The first one is the example of how to not write a solution. The second one is written by mueller.25, Geometrix and Amar_05, and tells us how to write a solution.
Let $ABCD$ be an isosceles trapezoid with $AB\parallel CD$. Let $E$ be the midpoint of $AC$. Denote by $\omega$ and $\Omega$ the circumcircles of the triangles $ABE$ and $CDE$, respectively. Let $P$ be the crossing point of the tangent to $\omega$ at $A$ with the tangent to $\Omega$ at $D$. Prove that $PE$ is tangent to $\Omega$.
Proof 1: Redefine $P$ as the pole of $DE$ wrt $\Omega$. We prove that $PA$ is tangent to $\omega$.
Let $G$ be the midpoint of $BD$ and $T$ be the intersection of $AC$ and $BD.$ Define $G= \omega \cap \Omega \neq E.$ (Note that $G$ is the reflection of $E$ across the perpendicular bisector of $AB$ and $CD$.) Also by applying Pascal on $(DDGEEC) \implies P_{\infty AB},P,T$ are collinear. Hence, $PT \parallel AB \parallel CD \parallel GE.$ And $\measuredangle DGE=\measuredangle DCE=\measuredangle PTE \measuredangle PET=\measuredangle PEC=\measuredangle EDC=\measuredangle DEG.$ And as found earlier, $\frac{PT}{TE}=\frac{GD}{GE}=\frac{BG}{GE} \implies PT= \frac{TE}{GE} \cdot BG=\frac{TA}{AB} \cdot BG\implies \frac{PT}{TA}=\frac{BG}{AB} \implies \triangle ATP \sim \triangle ABG$.
$\implies \triangle ATP $ similar $ \triangle ABG.$ So $\measuredangle PAE= \measuredangle PAT=\measuredangle GAB = \measuredangle ABE$
This solution is way more structured with the exact same content! Moreover, can you see how numbering the claims also made it easier to refer to? For example, we stated, "From claim 3."
Tip 2: Alignment and spacing
We will start with an example INMO 2005 P6. The first solution shows how to not write a solution and the second solution shows how to write a solution with perfect alignment. It must be perfect since your favorite blogger aka me has written it :D! ( If I am not your favorite blogger then 😁🗡)
Find all functions $f : \mathbb{R} \longrightarrow \mathbb{R}$ such that $$f(x^2 + yf(z)) = xf(x) + zf(y) , $$ for all $x, y, z \in \mathbb{R}$.
Solution1: Let $P(x,y)$ denote the assertion.Then$P(x,0,0)\implies f(x^2)=xf(x)\implies f(x)=-f(-x)$And$f(x^2+y^2)=f(x^2)+f(y^2)\implies f(a+b)=f(a)+f(b), a,b \in \mathbb{R}^{+}$Also $f(0)=0.$ Now we show that $f$ is injective at $0.$ So let $f(t)=0, t\ne 1.$ So$P(0,1,t)\implies 0=f(f(t))=tf(1)\implies f(1)=0\text{ or } f=0$Assuming $f$ is non constant, we get $f(1)=0.$ Then$P(0,a,1)\implies 0=f(a)\forall a.$Not possible, hence $f$ is injective at $0.$Moreover,$P(0,1,x)\implies f(f(x))=xf(1).$Note that $f(f(x))=xf(1)\implies f \text { is bijective }$Actually, we don't need the following claim, but I found it too good to not add.Note that $f(1)=1$ As, $P(0,y,1)\implies f(yf(1))=f(y)\implies yf(1)=y\implies f(1)=1.$ Another way is, Using surjectivity,$\exists k, f(k)=1\implies P(0,k,k)\implies 1=f(kf(k))=k\implies k=1$ Now, fix $x.$ We carry $y$ and choose $z$ such that $f(x^2+yf(z)=xf(x)+zf(y)=0.$Then$x^2+yf(z)=0, xf(x)+zf(y)=0.$ Take $y=x\implies z=-x$ in $xf(x)+zf(y)=0.$ So, we get$x^2+xf(-x)=0\implies f(-x)=-x\implies f(x)=x.$ So${ f=0,f(x)=x}$
Solution2: The solutions are $\boxed{ f=0,f(x)=x}$
Let $P(x,y)$ denote the assertion.
Then$$P(x,0,0)\implies f(x^2)=xf(x)\implies f(x)=-f(-x)$$
And$$f(x^2+y^2)=f(x^2)+f(y^2)\implies f(a+b)=f(a)+f(b), a,b \in \mathbb{R}^{+}$$
Also $f(0)=0.$ Now we show that $f$ is injective at $0.$ So let $f(t)=0, t\ne 1.$ So$$P(0,1,t)\implies 0=f(f(t))=tf(1)\implies f(1)=0\text{ or } f=0$$Assuming $f$ is non constant, we get $f(1)=0.$ Then$$P(0,a,1)\implies 0=f(a)\forall a.$$
Not possible, hence $f$ is injective at $0.$
Moreover,$$P(0,1,x)\implies f(f(x))=xf(1)$$
Note that$$f(f(x))=xf(1)\implies f \text { is bijective }$$
Actually, we don't need the following claim, but I found it too good to not add.
Claim: $f(1)=1$
Proof:$$P(0,y,1)\implies f(yf(1))=f(y)\implies yf(1)=y\implies f(1)=1$$
Alternate proof: Using surjectivity,$$\exists k, f(k)=1\implies P(0,k,k)\implies 1=f(kf(k))=k\implies k=1$$
Now, fix $x.$ We carry $y$ and choose $z$ such that$$f(x^2+yf(z)=xf(x)+zf(y)=0$$Then$$x^2+yf(z)=0, xf(x)+zf(y)=0$$
Take $y=x\implies z=-x$ in $xf(x)+zf(y)=0.$ So, we get$$x^2+xf(-x)=0\implies f(-x)=-x\implies f(x)=x$$
Hence the solutions are $\boxed{ f=0,f(x)=x}.$
The first solution and the second solution have exactly the same content! However, the second solution cares about spacing and alignment! Now, try reading both the solutions and write in the comments which one was easier to read :D
As stated in the article written by Ed Barbeau (linked below), "[split] your solution so that each section deals with a single main idea. It should be possible for the reader to glance over your solution and get a sense of its architecture, the main steps, and your methods of dealing with them."
Tip 3: Using asterisk or roman numerals
Many times, we need to refer to some part of the solution, and hence using numbers, asterisks or roman numerals is very useful!
For example, we show a solution written by Srijonrick.
Theorem: Let $ABC$ be a triangle and let $A_1$ be a point on the line $BC$. Then, prove that$$\dfrac{A_1B}{A_1C}=\dfrac{AB}{AC} \cdot \dfrac{\sin A_1AB}{\sin A_1AC}.$$
Proof: By sine rule in $\triangle AA_1B$ we get:$$\frac{A_1B}{\sin A_1AB}=\frac{AB}{\sin AA_1B}-(1)$$Again by sine rule in $\triangle AA_1C$ we get:$$\frac{A_1C}{\sin A_1AC}=\frac{AC}{\sin AA_1C}-(2)$$Since $\sin AA_1B=\sin (180^{\circ}-AA_1B)=\sin AA_1C$, hence on dividing $(1)$ by $(2)$ we get:$$\dfrac{A_1B}{A_1C}=\dfrac{AB}{AC} \cdot \dfrac{\sin A_1AB}{\sin A_1AC} \quad \blacksquare$$
We can see how he used the notations $(1), (2)$ and by introducing these two notations, it helped the reader see which two equations are we dividing, etc.
Tip 4: Case work
There isn't much to talk about this topic, however, many cases might have subcases! Hence, make sure to use meaningful notation to denote the cases and subcases and subcases's subcases. For example, what I do is,
Cases: Case 1, case 2, case 3,
Subcases: Case 1.1, case 1.2, case 1.3
Subcase's subcase: case 1.1.1, case 1.1.2
Here is an example! The solution is again provided by your favourite blogger aka me 😎
Swedish MO: Determine all positive integers $k$, $\ell$, $m$ and $n$, such that$$\frac{1}{k!}+\frac{1}{\ell!}+\frac{1}{m!} =\frac{1}{n!} $$
Solution:The only solution is $\boxed{(k,l,m,n)=(3,3,3,2)}$
Note that $n<k,l,m.$ Now we got cases.
Case1:$k-n=1$
Then$$\frac{3}{k!}=\frac{1}{k-1!}\implies k=3$$
Case2: $k-n=2$
Then we have three sub case:
Case 2.1:$$\frac{1}{k!}+\frac{1}{(k-1)!}+\frac{1}{k!}=\frac{1}{k-2!}\implies \frac{2}{k\cdot k-1}+\frac{k}{(k-1)k}=1\implies 2+k=k(k-1).$$We compute. There are no real solutions.
Case 2.2:$$\frac{1}{k!}+\frac{1}{k-1!}+\frac{1}{k-1!}=\frac{1}{k-2!}\implies \frac{1}{k\cdot k-1}+\frac{2k}{(k-1)k}=1\implies 1+2k=k(k-1)$$This is not possible as $k\cdot k-1$ is even.
Case 2.3:$$\frac{3}{k!}=\frac{1}{k-2!}$$It's not possible.
Now assume $k-n\ge 3.$ And also assume $m\le l\le k.$
Then we have$$1+\frac{k!}{l!}+\frac{k!}{m!}=\frac{k!}{n!}$$We have RHS even, so we have $l=k-1.$
But then we have $k|$RHS. So we have $m=k.$ Which is not possible because $m\le l.$
So the only solution is $\boxed{(k,l,m,n)=(3,3,3,2)}$
Tip 5: Handwriting
I cannot emphasize this enough! HANDWRITING DOES NOT MATTER. A myth is that handwriting matters in a contest and no, it never does! What matters is that it should be readable and neat.
Handwriting never ever directly matters to the marks but again. $$ \text{Solution is readable } \implies \text{ graders happy }$$ $$ \implies \text{ you get deserving marks } \implies \text { you happy } $$ Also here, readable handwriting doesn’t mean topper's handwriting!
In fact, I think using block letters is generally helpful (No, please don't try to change your handwriting, your goal is to make the reader understand your idea). I prefer blocks because in subjects that have their own syntax and a set of specific symbols that mean different things it becomes difficult to understand. For example, $\ell$ is similar to the cursive $l$...
Surely, I can read this but wth is this!!!!!! Like WTH!! Give some sympathy to the reader :( And please stop being lazy about your handwriting 😑
And if you do not want to change your handwriting, at least make sure to write neatly!!! We have another example by your favourite IMOTCer Atul!!!
Let's see another example of how to write a geometry solution ( again by your favourite math girl aka me)
Sharygin 2022 P16: Let $ABCD$ be a cyclic quadrilateral, $E= AC \cap BD , F = AD \cap BC .$ The bisectors of angles $AFB$ and $AEB$ meet $CD$ at points $X, Y.$ Prove that $A, B, X, Y$ are concyclic.
Solution: Define $K:=EX\cap AB, J:=FY\cap AB.$ Define $G:= AB\cap CD.$ Define $Q$ as the Miquel point of $ABCD.$ We begin with the following claim about $K-E-X$ and $F-J-Y.$
Claim: $K-E-X||F-J-Y.$
Proof: This is simply angle chase. Let $\angle BDC=\theta, \angle ADB=\alpha, \angle ACB=\alpha, \angle ACD=\beta.$
Note that $\angle EDX=90-(\theta+\beta)/2, \angle DFY=90-\alpha-(\theta+\beta)/2.$ And we get that$$\angle EXD=90+\beta/2-\theta/2=\angle FYD\implies FY||EX$$
We begin with the following lemma.
Lemma: $GX=GK$
Proof: Note that$$\Delta AKE\sim \Delta DXE \implies \angle GXK=\angle GKX$$
Now to prove that $AXYB$ is cyclic, by Power of Point, it's enough to show that$$GA\cdot GB=GD\cdot GC=GX\cdot GY.$$This motivates use to invert wrt $(G, GA\cdot GB).$
Let $K'$ and $X'$ be the inverted image of $K,X$ respectively. Note that after inversion, we will have$$\angle GXK=\angle GKX=\angle GX'K'\implies K'X'||KX.$$So it's enough to show that $K'-X'-F$ collinear and we will be done as by using our \claim, we get $K'=J,X'=Y.$
Define $Q$ as the Miquel point of $ABCD.$ Hence we have $QFBA$ cyclic and we also have $Q\in FG\implies GQ\cdot GF=GA\cdot GB\implies Q\text{ is the inverted image of }F.$
Now, to show that $F-K'-X',$ enough to show that $AGKX$ is cyclic.
But first, we prove the following lemma.
Lemma: $AK/KB=DX/XC$
Proof: Note that$$\frac{AK}{KB}\cdot \frac{XC}{DX}=\frac{AE}{EB}\cdot \frac{EC}{ED}=\frac{AE\cdot EC}{EB\cdot ED}=1$$
Using the well-known fact that there is a spiral similarity centred at $Q$ which takes $DC\leftrightarrow AB\implies \Delta AQB\sim \Delta DQC.$
Using the lemma and the similarity, we get that $\angle QKG=\angle QXG.$
This implies that$$ QKXG\text { is cyclic}.$$We are done!
In this example, we defined all the points first then we also added 1 diagram in the fair solution, but tbh while doing rough I had drawn 5+ diagrams :P.
Tip 7: How much detail to include
As Evan says here, "You lose points if a student who did NOT solve the problem could have written the same words as you. For example, whenever you say something like “it’s easy to see $X$”, the grader has to ask whether you actually understand why $X$ is true, or don’t know and are just bluffing. So that’s always the criteria you should have in your head when deciding what needs to be written out in full" So your goal is to convince the reader that you know the solution.
Also, I think it is completely okay to skip minute details that you know a grader would obviously know .. for example, As $\angle B=10,\angle C=80$ hence $\angle A=90$ (sum of interior angles of a triangle is 180).
There is literally no harm in writing a very detailed solution, the only problem is time. So try to write a solution that has a sufficient amount of details and is good enough to convince the grader that you know the solution. If you think there is some time left, after a write up then you can add minor details. There is no harm in adding them as it reduces your chances of getting -1s.
I think my most detailed write up was the solution to the Application problem of CAMP 2021 P1. Here's my solution to CAMP 2021 Application P1.
Too nice
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